1. IIT–JEE syllabus

Acids and Bases, pH, Common ion effect, Buffer solutions, Hydrolysis of salts and Solubility Product.   

2. Acids and Bases

Here we discuss some important definitions of acids and bases.

2.1  Arrhenius Definition 

Arrhenius defined an acid to be a substance which when dissolved in water, dissociates to give H+ ions. A base was defined as a substance which when dissolved  in water, dissociates to give OH–  ions.

2.2  Bronsted – Lowry Definition 

According to this definition an acid is a proton donor and a base is a proton acceptor. This definition makes no mention of the solvent (and is applicable even if no solvent is present).

3. pH of Acids and Bases 

Many properties of aqueous solutions depends on the concentration of H+ ions of the solutions and therefore there is a need to express these concentrations in simple terms. For this purpose we introduce the concept of pH. 

pH  = – log aH+ (Where aH+ is the activity of H+ ions).

Activity of H+ ions is the concentration of free H+ ions in a solution. By free, we mean those that are at a large distance from the other ion so as not to experience  its pull. We can infer from this that in dilute solutions, the activity of an ion is same as its concentration since more number of solvent molecules would separate the two ions. For concentrated solutions the activity would be much less than the concentration itself.

Therefore, the earlier given expression of pH can be modified for dilute solutions as,
pH  = – log [H+]. This assumption can only be made when the solution is very much dilute, i.e, [H+] ≤ 1M. For higher concentration of H+ ions, one needs to calculate the activity experimentally and then calculate the pH.

3.1  Strong Acids 

Let us now see how to calculate the pH of a solution of a strong acid in water (it should be noted that pH calculations are only made for aqueous solutions). Let the strong acid be HCl. If we take 10–1M HCl, the [H+] would be 10–1 M, as HCl is a strong acid and would dissociate completely. Therefore the pH would be, 

pH = –log 10–1 = 1

Concentration of HCl	pH
10–1M	1
10–2M	2
10–3 M	3
10–4 M	4
10–5 M	5
10–6 M	6
10–7 M	7 (?)

We can see that for 10–7M of HCl we have some hesitation in talking about the pH. This is because if we use our expression of pH, it works out to be 7 which is somehow associated with neutrality. We shall now explain how to calculate the pH of 10–7M HCl. Before we do this we shall discuss the dissociation of water.

3.2 Dissociation of Water and Ionic Product of Water 

Water behaves as a weak acid (or weak base) as it weakly dissociates to give H+  ions (or OH–).

H2O  H+ + OH–

This weak dissociation of water makes it reach equilibrium with its ions. 

∴ K =

The concentration of water before dissociation is 55.56 moles per liter. Since the dissociation of water is so feeble that it can be assumed that the concentration of water at equilibrium is almost the same. Therefore we can assume that [H2O]  is a constant .

∴ K[H2O] = [H+] [OH–] = KC

We can see that the above expression is KC. In this topic we shall give special sub–scripts for different types of KC ‘s. Hence the above expression’s which is the product of the concentrations of H+ and OH– is called the ionic product of water, represented as Kw. At 25°C the value of Kw is approximately equal to 10–14. 

∴ [H+] [OH–] = 10–14 …(1)

As water gives the same amount of H+ and OH– and if we represent the concentration of H+ as xM then the concentration of OH– will also be xM.

∴ x × x = 10–14 

 x2 = 10–14 …(2)

x = 10–7M …(3)

⇒ pH  = – log 10–7 = 7 

Now we revert back to our original problem which is the calculation of pH of a solution, 10–7 M HCl.

When we add 10–7 M HCl in water, the [H+] from HCl would be10–7 M. But pH is the negative logarithm of the total H+ ion concentration of a solution and not that part which comes from only HCl ( you might wonder why we have changed our stand. This will become clear in a short while). Therefore, we need to add the concentration of H+ ions coming from water also. This can be done in two ways.

Method – I

We already know the concentration of H+ ions from pure water. It is 10–7M. So all we need to do is to add the H+ ion concentration from water and that from HCl to get the total H+ ion concentration.

∴ [H+]T = [H+]HCl + [H+]water 

= 10–7M + 10–7M 

= 2 × 10–7 M

∴ pH = – log [H+]T

= – log 2 × 10–7 

= 6.6989

Though everything looks OK, there is a big mistake we have made. In Lesson 4 (Chemical Equilibrium) we have learnt in Le–Chatlier’s Principle that when concentration of a reactant or product is changed the reaction tends to go forward or reverse to finally reach equilibrium. We can see that  in this case by adding 10–7 M HCl  to water, we are increasing  the concentration of H+ ions. This should make the reaction go in the reverse direction. Therefore the above given method is incorrect.

Method – II

The statement given above that “the addition of an ion to an equilibrium, having the same ion makes the equilibrium reaction move in a direction to consume that ion” is called the common ion effect. This implies that water would dissociate less in the presence of HCl. Let the amount to which water disassociates be x′ in the presence of 10–7 M HCl. 

∴ H2O H+ + OH–

At equilibrium: x′ + 10–7 x′

 [H+] [OH–] = 10–14

⇒ (x′ + 10–7) (x′) = 10–14

Calculating for x′, we get x′ = 0.618 × 10–7 M.

One can clearly see the common ion effect in action. Water which was dissociating to give 10–7M H+ ions, has now experienced the common  ion effect and has finally yielded 0.618 × 10–7 M H+ ions.

Therefore, [H+]T = 10–7 + 0.618 × 10–7 

= 1.618 × 10–7 M

∴pH    = – log ( 1.618 × 10–7) =  6.7910

Now, lets answer the question as to why we take the H+ ions of water into account for calculating the pH of 10–7 M HCl while we never considered it for calculating the pH of
10–6M, 10–5M, 10–4M, 10–3M, 10–2M and 10–1M HCl. It can be seen that the H+ ions from water has decreased due to the common ion effect.  Greater the of the common ion added, greater will be the effect. Therefore for concentrations higher than 10–6M (and inclusive of 10–6M), the H+ from water will be even less than 0.618 × 10–7 M and would be so small in comparison to the [H+] from HCl, that we can ignore the contribution from it. So, finally we conclude that H+ ions of water needs to be considered only if an acid is present such that [H+]acid < 10–6M.

Illustration 1: 1 mL of 10–5 M HCl solution is diluted to 1000 mL. Calculate the pH of the resulting solution.

Solution: [H+] = [HCl] = 10–8M (after dilution)

Since this is lower than even the [H+] from dissociation of H2O so we must consider water as the principal source of H+ ion and HCl as exerting common ion effect on the dissociation of water.

H2O +

Kw = [H+] [OH–]

(10–8 + x)x = 10–14

(Kw = 10–14 at 25°C) ⇒ x2 + 10–8x – 10–14 = 0

x =  = 9.5125 × 10–8

[H+]= 9.5125 × 10–8  + 10–8 = 10.5125 × 10–8 

pH = 6.9783

The same logic applies for bases also. The only difference is that for deciding whether we should take the OH– ion from water or not, we should see whether [OH–] from the base is less than 10–6 M or not. 

3.3 Weak Acids 

Let us take a weak acid (CH3COOH) and see how to calculate its pH. Let its initial concentration be C moles / l.

CH3COOH CH3COO–     + H+

Initial C 0 0

At eq. C(1–α) Cα Cα

∴ KC =

This equilibrium constant,  KC is given a symbol Ka.

∴ Ka = =   =

Generally for weak acids, α (degree of dissociation) is very much less than 1. Therefore,

Ka = Cα2 ⇒ α =

We can see that as concentration decreases, α increases.

[H+] = Cα =

Illustration 2: Calculate pH of the following two solutions

1. i) 1 M CH3COOH, and
2. ii) 10–6M CH3COOH

Ka(CH3COOH) = 1.8 × 10–5

Solution: i) The concentration of CH3COOH being appreciable, α will be negligible in comparison to unity and hence we may use the approximation.

[H+] =  

⇒ pH = – log[H+] = – log (1.8 × 10–5]1/2 = 2.37

1. ii) We solve this problem by following two methods: one using approximation and another without it so as to see the difference.

Method 1: Since [H+] due to a weak acid is given by 

[H+] = = = 4.24 × 10–6 

pH = 5.37

Though we seemed to have solved it correctly, there is an error that we have made. This error can be highlighted by considering the pH of 10–6M HCl. We can see that the pH would be 6. Now, we know that lesser pH implies higher concentration of H+ ions. So how can a weak acid having the same concentration as a strong acid disassociate to give the same concentration of H+ ions.

The mistake has occurred in the assumption that α is very much small compared to 1. In fact by making this assumption the α that we calculate is, 

α = = = 4.24 

This value is not possible as α cannot be more than 1, ever. Therefore, we calculate the pH without this assumption.

Method 2: Ka = , 1.8 × 10–5 =

Solving for α, α = 0.95

∴ [H+] = Cα = 0.95 × 10–6 = 9.5 × 10–7 ⇒ pH = 6.02 

Of course we need to take the [H+] from water also as [H+]acid < 10–6M.

∴ (x′  + 9.5 × 10–7 ) x′ =  10–14 (Where x′ is the amount to which  water dissociates in the presence of 10–6M CH3COOH).

 x′ = 1.04 × 10–8

∴ [H+]T = 9.5 × 10–7 + 1.04 × 10–8

pH = 6.01 

This illustration shows a need to know as to when is the assumption that α is very small compared to 1, valid. This is done in the following  manner.

First of all calculate the value of α from  the expression  α =

If this value of α comes out to be less than or equal to 0.1, then the assumption is valid. If the value of α is such that, α>0.1, then the assumption is not valid and one has to calculate α using the quadratic expression. 

4. Hydrolysis of Salts 

Salts are the product of an acid and a base, other than water. Depending on the nature of an acid or a base there can be four types of salts:

1. i) Salt of a weak acid and a strong base,

1. Salt of a strong acid and a weak base,
2. Salt of a weak acid and a weak base and
3. Salt of a strong acid and a strong base.

We shall first look at what is hydrolysis and then find out how to calculate the pH due to it.

4.1 Salt of a Weak Acid and Strong Base  

Let us take a certain amount of weak acid (CH3COOH) and add to it the same amount (equivalents) of a strong base (NaOH). They will react to produce CH3COONa. 

CH3COOH + NaOH ⎯→ CH3COONa + H2O

CH3COONa being a strong electrolyte, completely dissociates into its constituent ions.

CH3COONa ⎯→ CH3COO– + Na+

Now, the  ions produced would react with H2O. This process is called hydrolysis. 

Na+ + CH3COO– + H2O   CH3COOH + NaOH

We know that NaOH is a strong base and therefore it would be completely dissociated to give Na+ and OH–  ions.

∴ Na+ + CH3COO– + H2O CH3COOH + OH– + Na+ 

Canceling Na+ on both the sides, 

⇒ CH3COO– + H2O   CH3COOH + OH–

We can note here that ions coming from strong bases do not get hydrolysed. We should note here that the solution will be basic. This is because the amount of CH3COOH produced and OH-– produced are equal. But CH3COOH will not completely dissociate to give H+ ions. Therefore [OH–] ions  will be greater than [H+] ions.

Since the reaction is at equilibrium,

KC =   

This equilibrium constant Kc is given a new symbol, Kh.

∴

If  we multiply and divide the above equation by [H+] of the solution, then 

Kh = =

• Kh = ⇒ Kh = =

CH3COO–+H2O CH3COOH  + OH– 

Initial: C   0   0

At eq: C(1–α)   Cα Cα

Where α is the degree of hydrolysis of CH3COO– ion. 

∴ =

If α is very much less than 1, 

Cα2 = ,  α =

As [OH–] = Cα, [OH–] = C ×

[H+] =

or pH = – log [H+] = –log =  

Now, we will analyse the hydrolysis of this salt in detail. The reactions occurring were, 

CH3COOH + NaOH ⎯→  CH3COONa +  H2O …(1) 

CH3COONa ⎯→ CH3COO– + Na+ …(2)

CH3COO– + H2O CH3COOH + OH– …(3)

If we look at the first reaction, we can see that NaOH will be fully ionized as Na+ & OH– ions and CH3COONa as CH3COO– & Na+ ions . Canceling Na+ from both the sides, the first reaction looks like,

CH3COOH + OH– ⎯→ CH3COO– + H2O … (4)

You can see clearly that reaction  (4) and (3)  are the reverse of each other. Therefore, reaction (4) is also in equilibrium.

∴ CH3COOH + OH– CH3COO– + H2O … (5)

This actually puts us in a dilemma. If we look at reaction (5) [which is basically the reverse of reaction (1)], we see that when CH3COOH reacts with OH– the reaction reaches equilibrium. Now, there are two possibilities: (i) either acetic acid reacts with OH– and the reaction reaches equilibrium or (ii) acetic acid reacts with OH– and the reaction completely goes to produced CH3COO– which then gets hydrolyses (by going in the reverse direction) and finally reaches equilibrium. On close scrutiny we would realize that the first possibility is more realistic. It is not logical for a reaction to completely go to the right and then go reverse to reach equilibrium. Now, if the first explanation is correct then we find that actually no hydrolysis has taken place. What has happened is that the first reaction has not gone to completion and has reached equilibrium.

Apart from this difference, both the explanation would give us the same result in terms of pH calculation. For example, we can see that if a certain amount of CH3COOH reacts with the same amount (equivalents) of OH– ions and the reaction reaches equilibrium, then equal amounts (equivalents) of OH– ions and CH3COOH would be left over. This is a similar situation which we encountered earlier in hydrolysis. We can see that this solution would be basic.

To conclude we can say that hydrolysis of a salt occurs only when the salt is dissolved in water.

Illustration 3: 100 mL 0.1 M CH3COOH is titrated with 0.1 NaOH solution. Calculate the pH of the solution at the equivalence point. Given pKb (CH3COO–) = 9.26.

Solution: The equivalence point will reach when 100 mL 0.1 M NaOH solution is added

100 mL 0.1 M CH3COOH ≡ 10 milli mole CH3COOH

100 mL 0.1 M NaOH ≡ 10 millimole NaOH

10 millimole CH3COOH will react with 10 millimole NaOH to form 10 millimole salt (CH3COONa).

CH3COOH + NaOH ⎯→ CH3COONa + H2O

10 millimole 10 millimole 0 (initially)

0 0 10 millimole (after completion of the reaction)

Molarity of salt solution = = = 0.05

The anion of the salt being a conjugate base of weak acid, the same must be very strong and will undergo hydrolysis to furnish free OH– ions in solution according to the equation.

CH3COO– + H2O CH3COOH + OH–

The pOH of the resulting alkaline solution can be given as

pOH = [pKw – pKa – logC]

= [14 – 4.74 – log0.05] = 5.28

∴ pH = 14 – 5.28 = 8.72

Note that pKa + pKb = pKw, so pKa = pKw – pKb = 14 – 9.26 = 4.74

4.2 Salt of a Weak Base and a Strong Acid 

Let the acid be HCl and the base be NH4OH.

Therefore the salt would be NH4Cl.

NH4Cl completely dissociates into  and Cl– ions. 

∴NH4Cl ⎯→ + Cl–

Cl– + +H2O NH4OH + HCl

HCl being a strong acid dissociates completely to give H+ ions and Cl– ions. 

∴ Cl– + + H2O NH4OH + H+ + Cl–

+ H2O  NH4OH + H+

We can see that the ion coming from the strong acid does not get hydrolysed. We had previously seen that the ion coming from the strong base also does not get hydrolysed. Hence, one can conclude that the salt of a strong acid and weak base does not get hydrolysed.

In this hydrolysis, NH4OH and H+ are being produced. This implies that the solution is acidic. To calculate pH,

+ H2O NH4OH   + H+

Initial: C 0 0

At eqb: C(1–α)     Cα Cα

Where α is the degree of hydrolysis of .

Kh =

Multiplying and dividing by OH– and rearranging,

Kh =

⇒  = =  

∴ Kh =

Now, substituting the concentrations, 

Kh =

If α ≤ 0.1, then, Cα2 =

α =

Since [H+] = Cα, [H+] = C = or pH =

4.3 Salt of a Weak Acid and Weak Base

Let the weak acid be CH3COOH and the weak base be NH4OH. Therefore, the salt is CH3COONH4.

The salt completely dissociates. 

CH3COONH4 ⎯→ CH3COO– + 

The ions get hydrolysed according to the reaction. 

CH3COO– + + H2O  NH4OH + CH3COOH

Initial : C C 0 0

At equilibrium: C(1–α) C(1–α) Cα Cα

Kh =

Multiplying and dividing by H+  & OH– and rearranging, 

Kh =  ⇒  ⇒ 

∴ Kh =

Substituting the concentration terms,

Kh = ∴

There is an important issue that needs clarification before we move on further. In this case, we can see that both the ions (i.e, cation and anion) get hydrolysed to produce a weak acid and a weak base (hence, we can’t predict whether the solution is acidic, basic or neutral). We have considered the degree of hydrolysis of both the ions to be the same. Now we present an explanation as to why this is incorrect and then state reasons for the validity of this assumption.

Actually the hydrolysis reaction given earlier, 

CH3COO– + + H2O  CH3COOH + NH4OH

is made up of the following three reactions,

CH3COO– + H2O  CH3COOH + OH–

+ H2O  NH4OH + H+

H+ + OH–  H2O

If we add these three reactions, the net reaction is the one stated previously. This suggests that both CH3COO– and get hydrolysed independently and their hydrolysis depends on: (i) Their initial concentration and (ii) the value of Kh which is for CH3COO– and for . Since both of the ions were produced from the same salt, their initial concentrations are same. Therefore unless and until the value of and or Ka  and Kb is same, the degree of hydrolysis of the ions cannot be same.

To explain why this assumption is valid, we need to now look at the third reaction, i.e., combination of H+ and OH– ions. It is obvious that this reaction happens only because one reaction produced H+ ion and the other produced OH– ions. We can also note that this reaction causes both the hydrolysis reaction to occur more since their product ions are being consumed. If you look at the solution of Exercise 1, you will notice that the equilibrium which has smaller value of equilibrium constant is affected more by  the  common ion effect. For the same  reason if for any reason a reaction is made to occur to a greater extent by the consumption of any  one product ion, the reaction with the  smaller value of equilibrium constant tends to get affected more.

Therefore, we conclude that firstly the hydrolysis of both the ions occurs more in the presence of each other (due to the consumption of the product ions) than in each other’s absence. Secondly the hydrolysis of the ion which occurs to a lesser extent (due to smaller value of Kh) is affected more than the one whose Kh is greater. Hence  we can see that the degree of hydrolysis of both the ions would be close to each other when they are getting hydrolysed in the presence of each other. 

Now, to calculate the pH of this solution, we need to understand  one more principle. Let us assume that we have an aqueous solution of CH3COOH. Let us also assume that we do not know whether CH3COO– and H+ are in the solution or not (even though H+ would be there since the solution is aqueous). We can see that the Qa for the reaction

CH3COOH  CH3COO– + H+ is zero (if we assume no CH3COO– or  H+). 

∴ Qa < Ka. The reaction would therefore move forward to reach equilibrium. The same is true when CH3COO– and H+ are present without CH3COOH. Hence, we  conclude that “Any equilibrium that can exist (i.e., when all reactants and / or all products are present) will exist (except when pure solids or liquids  are not present, in which case this statement will not be true)”.

As can be seen in the hydrolysis reaction, 

CH3COO– + + H2O NH4OH + CH3COOH,

CH3COOH is present in the solution. This implies that the equilibrium between CH3COOH, CH3COO– and H+ can exist and therefore would exist.

∴CH3COOH CH3COO– + H+

In fact the equilibrium between NH4OH, and OH– also exists. 

Now, we calculate the pH of the solution as, 

CH3COOH  CH3COO– + H+

Cα   C(1–α)

Ka = = ∴ [H+] = Ka ×

Substituting as

[H+] = Ka× = Ka × =

or pH =

5. Buffer Solutions 

A buffer solution is a solution which resists a change in its pH when such a change is caused by the addition of a small amount of acid or base. This does not mean that the pH of the buffer solution does not change (we make this assumption while doing numerical problems). It only means that the change in pH would be less than the pH that would have changed for a solution that is not a buffer.

There are three types of buffer solutions: 

1. i) weak acid–salt buffer 
2. ii) weak base–salt buffer and 

iii) salt buffer 

5.1 Buffer of a Weak Acid and its Salt with a Strong Base 

It is possible  to prepare a buffer solution by the addition of a weak acid and a salt of the acid with a strong base.

We shall explain the buffer action by the following example. Let us consider a buffer solution made up of CH3COOH and CH3COONa. The weak acid dissociates to a very small extent more so due to the common ion effect of its salt.

CH3COOH  CH3COO– + H+

CH3COONa ⎯→ CH3COO– + Na+

Now let us assume that a buffer solution contains 20 moles of CH3COOH and 20 moles of CH3COONa. The salt being a strong electrolyte would completely dissociate while the acid would be hardly dissociated. We assume that the amount of CH3COO– is 20 moles as the contribution from CH3COOH would negligible. We also assume that the amount of CH3COOH to be 20 moles as it would be very weakly dissociated. Therefore the solution contains 20 moles of CH3COOH and 20 moles of CH3COO– ions.

Let us now add 10 moles of H+ ions  to this solution. These 10 moles of H+ ions  would react with the 20 moles of CH3COO– ions to produce 10 moles of CH3COOH. (This is because the reaction  of CH3COOH  to give CH3COO– ions and H+ ions has an equilibrium constant value of approximately 10–5. Therefore, the reverse reaction, that is the reaction of CH3COO– ions  to combine with H+ ions to give CH3COOH would have an equilibrium constant of approximately 105. So this reaction can be assumed to be complete). These 10 moles of CH3COOH formed would ionize weakly because it is a weak acid. Moreover due to the common ion effect of CH3COO– ion (10 moles, left over), it would ionize even less. So the amount of H+ ion produced back is much less than the 10 moles that were added. So effectively, 10 moles of H+ ions were consumed and an amount much less than that is produced back which causes an insignificant change in pH of the buffer solution.

Let us add 10 moles of OH– ions to the same buffer solution containing 20 moles of CH3COOH and 20 moles of CH3COO– ions. These 10 moles of OH– ions added would be consumed by the 10 moles of CH3COOH to produce 10 moles of CH3COO– ions
(CH3COOH + OH– ⎯→ CH3COO– + H2O. This is because the reverse reaction is the hydrolysis of CH3COO– ions which has an equilibrium constant of ≈ 10–9.
So the forward reaction’s equilibrium constant would be  109, which  implies that the reaction is practically complete). The 10 moles of CH3COO– ions produced would be hydrolysed weakly (due to a very low K value of 10–9). On top of that, the presence of 10 moles of CH3COOH would further hamper the hydrolysis process and thereby the
OH– ion produced is much less than the amount that was added. This causes a
minor change in pH.

For a buffer to act as a good buffer the amount of CH3COOH and CH3COO– ion should be high. 

The pH of a buffer can be calculated as follows. 

Since acetic acid is in equilibrium,

= Ka

⇒ = Ka

⇒ [H+] = Ka 

⇒ log [H+] =  log Ka + log 

– log [H+] = – log Ka + log

⇒ pH = pKa  + log

This equation is called  Henderson – Hasselbalch equation. 

 

When a weak dibasic acid is taken then the pH at half neutralisation is given by a special formula which is derived by balancing charge and mass of reactants and products. For example, at half neutralisation of H2CO3 by NaOH, the pH is given by

pH =

Where and are 1st and 2nd dissociation constant of H2CO3.

5.2 Buffer of a Weak Base  and its Salt with a  Strong Acid 

This combination works on the principle as stated in the previous section. If we take the example of NH4OH and NH4Cl, then the OH– ion added would react with (from the salt) to produce NH4OH. The H+ ion added would react with NH4OH to produce .

The Henderson – Hasselbalch equation appears as

pOH = pKb + log

Exercise 1: a) Calculate the pH after the addition of 90 ml and 100ml respectively of 0.1N NaOH to 100ml 0.1N CH3COOH (Given pKa for CH3COOH = 4.74)

1. b) What is pH of 1M CH3COOH solution? To what volume must one litre of this solution be diluted so that the pH of resulting solution will be twice the original value. Given : Ka = 1.8 × 10–5

Illustration 4: 100 mL 1 M CH3COOH is treated with 0.8 gm NaOH. Calculate pH of solution. How much NaOH needs to be added further so as to increase its pH to 4.74? Will the dilution of the mixture will change the pH of solution

(Ka = 1.8 × 10–5).

Solution: 100 mL 1M CH3COOH ≡ 100 × 1 i.e. 100 millimoles of CH3COOH 

0.8 gm i.e. 800 mg NaOH ≡ i.e. 20 millimoles of NaOH 

20 millimoles NaOH will react with 20 millimoles CH3COOH leaving behind 80 millimoles of free CH3COOH. The mixture will be thus, an acid buffer mixture whose pH can be calculated using the equation.

pH = pKa + log

= – log (1.8 × 10–5) + log

= 4.74 – 0.60

= 4.14

To make the pH of solution equal to 4.74, must be equal to 1. The acid present in the original acid solution is 100 millimoles. So we have to add 50 millimoles i.e. (=2g) of NaOH. Thus the amount of NaOH to be added further is 2 – 0.8 i.e. 1.2g.

Upon dilution the ratio and also pKa will remain unchanged and hence pH of solution will remain unchanged.

5.3 Indicators 

In the illustration 3, we have  calculated the pH during the titration of a weak acid with a strong base at various stages. If we calculate the pH at several instances during the titration and then plot a graph between the pH and the volume of strong base, we get a curve as shown in figure 1.

The curve has been drawn by considering the titration of 50 ml of 0.1M CH3COOH with 0.1M NaOH.

Let us first of all understand the nature of the curve. Initially the pH changes rapidly with the volume of the base and the its change becomes very slow and again it changes rapidly. This phenomenon can be explained like this.  Initially, a strong base is added to a weak acid. Therefore the pH changes accordingly. But after the addition of a certain amount of the base, the solution turns into a buffer as contains CH3COOH and CH3COONa. So any further addition of NaOH causes a very little change in pH. Therefore the slow change.  As more and more of strong base is added, the salt concentration increases and the acid concentration decreases and especially near the equivalence point, the acid concentration is very small. When there is such a large difference in their concentration, the solution stops acting as a buffer and there is a sudden jump in the pH.

Now, let us assume that when 49.9 ml of NaOH has been added the pH of the solution is 6 ( this is just an assumption to illustrate the  role of indicators and not the actual value) and when 50 ml of NaOH is added (equivalence point) the pH is 8.69. If  we assume that one drop is equal to 0.1 ml then what we have is a situation in which on adding the last drop of NaOH the pH changes from 6 to 8.69.

An ideal indicator would be the one which changes its colour on the addition of the last drop of NaOH. Let the indicator be a weak acid HA (all acid base indicators are either weak acids or weak bases). It would dissociate weakly as, HA  H+ + A–. Let the weak acid have a colour red in acid solution and have a colour blue in basic solution. This implies that HA is red  (as HA would be almost undissociated in acidic solution) and A– is blue (as HA would be almost completely  dissociated in basic solution). Now,

Kind = where Kind is the Ka of the indicator. 

⇒ [H+] = Kind … (1)

To see any colour, we need to have a certain minimum concentration of the coloured species. Let the minimum amount of HA required to be present to make the colour red visible be 90% of the total indicator. This means that if the indicator has HA, 90% or more in amount, then we would see the colour red and if it is less than 90% then coloured red would not be visible.

Similarly we can assume that to see the colour blue, the minimum amount of A– required to be present is also 90%. This means that if 90% or more of the total indicator is A–, then colour blue would be visible, otherwise not.

We can see that the minimum ratio of to see the red colour would be  9. Let the [H+] at  this  instant (from equation 1) be 10–6.5. This is therefore, the minimum amount of H+ required to see the red colour. Converting this to pH (–log [H+]), we get the maximum pH  to see the red colour (i.e., 6.5). Similarly, the maximum ratio of to see the blue colour would be   . Let the [H+] at this instant (from equation 1) be 10–8. This is therefore the maximum amount of H+ required  to see the blue colour. Converting this to pH (–log [H+]), we get the minimum pH to see the blue colour (i.e., 8). Hence the indicator would change colour between pH 6.5 –8. It would be red at pH 6.5 or below. It would be blue at pH  8 or above. Between the two pH values, it would be colourless.

An indicator would be effective for a given titration if its range of pH for colour change (which in this case is between 6.5 to 8) is within the range of pH before and after the last drop of an acid or a base is added (which is in this case is between 6 to 8.69).

We can easily see that if this particular indicators is added in the solution, it would have red colour before the addition of the last drop and blue colour after the addition of the last drop.

5.4 Salt  Buffer 

The fundamental principle behind a buffer action is the fact that on adding an acid the system consumes the H+ ion added to produce a weak acid and on adding a  base, it consume the OH– ion added to produce a weak base. This ensures that H+ or OH– ion  added is consumed and the weak acid or the weak base produced gives less H+ or OH– ion as they are weak.

Based on this principle, a solution of a salt of a weak acid and weak base is also called as a buffer. Let us take the example of CH3COONH4. It dissociates as,
CH3COONH4 ⎯→ CH3COO– + . When H+ ion is added, CH3COO– ion  consumes it to give CH3COOH. When OH– ion is added, ion consumes it to give NH4OH. Hence it acts as a buffer.

[H+] = (derived earlier).

6. Solubility and Solubility Product  

Here we discuss the solubilities of ionic solutes. When a salt is dissolved in water, it dissolves as the salt, AgCl (s) ⎯→ AgCl(aq) and then the dissolved salt dissociates to give the ions AgCl (aq) ⎯→ Ag+ (aq) + Cl–(aq). After a certain amount of the salt has been dissolved, the solution would become saturated with the salt. Now, if some more of the salt is dissolved, the  salt would dissolve to give ions while at the same time the ions would precipitate to give back the salt. This amount (moles) of the salt that has made the solution saturated per liter of solution is called the solubility of the salt.

AgCl (s) Ag+(aq) + Cl– (aq)

∴K  =

Since the concentration of AgCl, which is a solid, is constant, 

⇒ K[AgCl] = [Ag+] [Cl–] = KC

This KC is given the symbol, KSP. This means that a solution cannot have the product of  the concentration of Ag+ and Cl– to be more than KSP (called solubility product).

6.1 Calculation of Solubilities of Salts 

We shall now discuss the  solubilities of different types of  salts under various conditions.

1. i) Solubilities of AgCl (salt of a strong acid and strong base) in water: 

AgCl would dissolve in water as, 

AgCl (s)  ⎯→ Ag+ (aq) + Cl– (aq)

At saturation point,

AgCl (s) Ag+(aq) + Cl– (aq)

If the solubility of the salt is x moles / l

∴ [Ag+] = xM, [Cl–] = xM ⇒ x2 =Ksp

x =

1. ii) Solubility of AgCl in a solution that is having 0.1M in AgNO3:

AgCl would dissolve and finally reach saturation. 

AgCl(s) Ag+(aq) + Cl– (aq)

The Ksp of AgCl is approximately 10–10 . If AgCl were to be dissolved in water (pure), its solubility would have been 10–5M (previous section). In the presence of 0.1M AgNO3 its solubility will decreases due to common ion effect. This means that [Ag+] from AgCl would be less than 10–5 M. Hence, we can ignore the contribution
of Ag+ from AgCl.

If the solubility of AgCl is x ′ moles / l in the presence of 0.1M AgNO3, then

[Ag+] = 0.1M , [Cl–] = x ′ M

∴ x′ = = 10–9 moles / l

iii) Solubility of CH3COOAg (salt of weak acid and strong base) in water: 

CH3COOAg dissolves and reaches saturation. Since it is a salt of weak acid and strong base, it would hydrolyse. If the solubility of the salt is x moles/ l then

CH3COOAg (s)  CH3COO– (aq)     + Ag+ (aq)

At eqb: x –y x

CH3COO– (aq) + H2O  CH3COOH (aq) + H+ (aq)

At eqb: x –y y         y

Where y is the amount of CH3COO– ion that is hydrolysed. 

∴ (x –y) x = Ksp

=

Knowing the values of Ksp and Ka, solubility of the salt can be calculated. 

1. iv) Solubility of CH3COOAg (salt of a weak acid and strong base) in an acid buffer of  pH = 4 (assuming that the buffer does not have any common ion by CH3COOAg):

CH3COOAg would dissolve and reach equilibrium. It would then be hydrolysed. If the solubility of the salt is x ′ M in this solution, then 

At eq.; CH3COOAg (s) CH3COO– (aq) + Ag+ (aq)

x′ –y′ x ′

CH3COO– (aq) + H+ CH3COOH (aq)  

x ′– y′   10–4 y′  

Since the solution is a buffer, the pH will be maintained. 

∴( x ′ – y′) x ′ = Ksp

=

Since in presence of basic buffer, the degree of hydrolysis will be suppressed by already existing –OH ions, therefore the approximated formula which can be used is

X′2 = (CH3COOAg) (neglecting y′′)

Knowing Ksp and Ka, the solubility can be calculated. 

1. v) Solubility of CH3COOAg in an buffer solution of pH = 9:

Following the same logic as give in the earlier section,

CH3COOAg (s)  CH3COO– (aq) + Ag+ (aq)

At eqb: x ′′–y′′     x ′′

CH3COO–(aq) + H2O  CH3COOH +  OH–

At eqb: x ′′ – y′′   y′′ 10–5

Where x′′ ′M is the solubility of the salt and y′′ the extent to which it is hydrolysed. 

∴( x ′′ – y′′) x ′′ = Ksp

=

Knowing, Ksp and Ka, the solubility can be calculated. 

3. vi) Solubility of AgCl in an aqueous solution containing NH3.

Let the amount of NH3 initially be ‘a’ M. If the solubility of the salt is x moles/ l, then 

AgCl (s) Ag+ (aq)  +  Cl– (aq)

At eq: x –y   x

Ag+(aq) + 2NH3 (aq)  Ag(NH3)2+(aq)

x–y   a–2y y 

Where y is the amount of Ag+ which has reacted with NH3.

∴( x –y) x = Ksp

= Kf ( formation constant of Ag(NH3)2+)

Knowing Ksp and Kf , the solubility  can be calculated. 

6.2 Precipitation of salts

For a salt (sparingly soluble) when dissolved in water : AB A+ + B– at equilibrium (saturation) Ksp = [A+] [B–]

When, we mix ions or if there be two or more ions in water, we define reaction coefficient (Q), called as a ionic – product (IP) in this case, giving the products of ions in water (ions of soluble salts and other common ions).

I.P. is product of ionic concentration due to ions already present in water or from a salt. I.P. may be and may not be equal to Ksp. To illustrate it more clearly , consider a case when 500 ml of an 0.005 M solution of AgNO3 is added to 500 ml of 0.001 M solution KCl. Now in solution (mixture), there are Ag+, , K+  and Cl– ions. The concentration of [Ag+] = = 0.005 = 0.0025 M (equal volumes are mixed) and [K+] = [Cl–] =
= 0.0005 M as equal volumes of two solutions are mixed.

Now, we know that Ag+ will react with Cl– ions to form AgCl since:

AgCl Ag+ + Cl– is a reversible reaction with a higher tendency towards left (solidifying or precipitating).

Now, question is, whether AgCl will be formed or not (precipitation of Ag+ and Cl– as AgCl) and if it formed, how much of it will be formed? For this we define some rules .

1. If ionic Product (IP) > Ksp ; precipitation takes place till I.P. equals Ksp
2. If Ionic Product < Ksp ; a precipitate will not be formed and the solution will be unsaturated 
3. If Ionic Product = Ksp ; a precipitate will not form an the solution is satuated in that salt. (or we can say that solution is at a critical stage, when precipitation just begins, but actually has not occurred yet in real sense).

In present case,

I.P. = [Ag+] [Cl–] = (0.0025) (0.0005) = 1.25 × 10–6

(Only for the salt which is sparing soluble not for KNO3)

Ionic Product > Ksp in this case (Ksp AgCl = 1.56 × 10–10)

Which mean precipitation takes place.

Exercise 2: a) The solubility of CaF2 in water at 20°C is 15.6 mg per dm3. Calculate the solubility product of CaF2.

2. b) The solubility of BaSO4 in water is 2.3 × 10–4 gm/100 mL. Calculate the percentage loss in weight when 0.2 gm of BaSO4 is washed with (a) 1lt of water (b) 1lt of 0.01 NH2SO4.

Illustration 5: 20 mL 0.1 M solution of a dibasic acid H2XO3 (K1 = 1 × 10–6) and K2 =1 × 10–11) is titrated with 0.2 M NaOH solution. Calculate pH of solution when (a) 5 mL and (b) 20 mL alkali had been added.

Solution: 20 mL 0.1 M H2XO3 ≡ 2 millimole of H2XO3

1. a) When 5 mL 0.2 M NaOH i.e. 1 millimole of NaOH has been added, 1millimole of NaHXO3 will be formed and 1 millimole of H2XO3 will remain free.

H2XO3 + NaOH ⎯→ NaHXO3 + H2O

2 millimole 1 millimole 0 initially

1 millimole 0 10 millimole when reaction is complete.

The resulting mixture will be an acid buffer mixture whose pH is as calculated below:

pH = pK1 + = 6 + = 6

1. b) 10 mL 0.2 M NaOH ≡ 2 millimole

2 millimole NaOH will react with 2 millimole H2XO3 to give 2 millimole NaHXO3 ( an acid salt). It is a point of half neutralisation.

H2XO3 + NaOH ⎯→ NaHXO3 + H2O

2 millimole 2 millimole 0

0 0 2 millimole 

The resulting solution will be the solution of NaHXO3. HXO3– is an amphiprotic anion i.e. it furnishes H+ ion in solution as an acid and HXO3– ion being a conjugate base of weak acid is a strong base which gets hydrolysed to give OH– ion in solution as

HXO3– H+ + XO32–

HXO3–+ H2O H2XO3 + OH–

Na+ ion is unhydrolysable. The pH of such a salt solution may be calculated using the equation

pH = = 8.5

1. c) 20 mL 0.2 M NaOH ≡ 4 millimole

4 millimole NaOH will react with 2 millimole H2XO3 to give 2 millimole Na2XO3. Thus it is the equivalence point of the titration.

XO32– being the conjugate base of a very-very weak acid HXO3– ion is a very-very strong base and in aqueous solution it would get hydrolysed to give free OH– ions in solution.

XO32– + H2O HXO3– + OH–, Kh =

HXO3– being a conjugate base of weak acid H2XO3 is also a strong base and will hydrolyse as

HXO3– + H2O H2XO3 + OH–, Kh = = 10–8

The hydrolysis constant XO32– ion being for more greater than that of HXO3–, it is wiseful to neglect the hydrolysis of HXO3– so as to simplify the calculation. The pOH of the resulting solution can be calculated using the formula of hydrolysis.

pOH = [pKw – pK2 – logC]

Molarity of Na2XO3 solution (C) = = 0.05

∴ pOH = [14 – 11 – log 0.05] = [14 – 11 + 1.30] = 2.15

∴ pH = 14 – pOH = 14 – 2.15 = 11.85

Illustration 6: The solubility of the sulphide of a trivalent metal in water at 25°C is 20.0 mg per dm3. Calculate its solubility in 0.5 M solution of Na2S assuming complete dissociation of Na2S. (M = 52 and S = 32)

Solution: Molecular formula of metal sulphide = M2S3

Molecular weight of metal sulphide = 52 × 2 + 32 × 3 = 200

Solubility in mole per dm3

= = 10–4 

M2S3(s)  2M3+  + 3S2–

2 × 10–4 3 × 10–4

Ksp = [M3+]2 [S–2]3 = (2 × 10–4)2 (3 × 10–4)3 = 1.08 × 10–18

0.5 M Na2S being completely dissociated will give 0.5 M S2– in solution. The S2– ion will exert common ion effect on the solubility of M2S3 and hence the solubility of M2S3 will decrease.

Let the solubility of M2S3 in 0.5 M Na2S is S mole per dm3

M2S3(s) 2M3+ + 3S2–

2S 3S + 0.5

Ksp will remain unchanged

(3S + 0.5)3 (2S)2 = 1.08 × 10–18

S being very small in comparison to 0.5, the above equation may be simplified as

(0.5)3 (2S)2 = 1.08 × 10–18

S = = 1.47 × 10–9 mole per dm3

7. pH of Mixtures of Acids & Bases 

Let us take x millimoles of acid (HA) and y millimoles of base (BOH). Note that acid is monobasic and base is monoacidic.

1. a) Strong acid & Strong base

• If x = y; then complete neutralization takes place and we get x(=y) millimoles of salt (BA) of strong acid  and strong base which means no hydrolysis takes place and pH of solution  = 7.
• If x > y; then there is an excess of strong acid and resulting solution is acidic with millimoles of acid left in excess  = x – y. 

Now if V c.c. be the volume of mixture, then:

⇒ M = . Now calculate pH using the equation pH = – log [H+]

• If x < y; then there excess of strong base and resulting solution is basic with millimoles of base  = y – x.

1. b) Strong base & weak acid

• If x = y; first of all neutralization takes place to give x(=y) millimoles of salt (BA). The salt will now undergo hydrolysis to give an alkaline solution. Calculate pH by using standard result:

pH  = 7 + (pKa + log C); C is concentration expressed in M  (mol/lt)

• If x > y;  there is excess of weak acid whose  millimoles  = x – y and y millimoles of salt is formed. This will give an acidic buffer solution. Calculate pH of buffer solution using Henderson’s Equation.

PH = pKa + log + log

• If x < y; the solution in this case contains excess of strong base whose millimoles are y – x.

M = . Calculate pH.

1. c) Strong acid & Weak base

• If x = y; first of all complete neutralisation takes place to produce x(=y) millimoles of salt (BA). The salt (BA) is of strong acid and weak base, hence hydrolysis takes place to give an acidic solution. Calculate its pH by using standard result.

pH  = 7 – (pKb + log C) ; C: is concentration of salt.

• If x > y; then solution contain excess of strong acid whose millimoles  = x – y.

M = . Calculate pH.

• If x < y; then there is an excess of weak base whose millimoles are  y – x and millimoles of salt (BA) are x. This will give a basic buffer solution. Calculate the pH by using Henderson’s Equation.

pH  = 14 – pKb – log = 14 – pKb – log

1. d) Weak acid & weak base

• If x = y; neutalisation takes place completely with the formation of x (=y) millimloles of salt (BA) of weak acid and weak base. So hydrolysis takes place. Calculate the pH by using: 

pH = 7 + (pKa – pKb)

• If x >y; then excess of weak acid (x – y) will remain with y millimoles of salt. This will give an salt buffer with acidic pH.
• If x <y; then excess of weak acid (y-x) and x millimoles of salt. This will give a salt buffer with basic pH.

 

8. Solution to Exercises

 

Exercise 1: a) If 90 ml. Of .1N NaOH is added to 100 ml. Of .1N CH3COOH, acidic buffer will form as 

H3CCOOH + NaOH  ⎯⎯→ H3COONa + H2

t = 0 .01 eq. .009 eq. 0 0

.001 eq. 0 .009 eq.

pH = pKa + log

= 4.74 + log = 4.74 + .9542 = 5.694

If 100 ml of 0.1N NaOH is added to 100 ml of 0.1N CH3COOH, complete neutralisation takes place and the concentration of H3COONa 

= M = .05M

Now, pH = 7 – pKa – log C

pH = 5.28

1. b) H3COOH + H2O H3COO– + H3O+

t = 0 1M 0 0

-xM xM xM

——————————————————————————

t = teq (1-x)M x x

Ka = ≈

∴ x = = 4.2 × 10–3 = [H3O+]

pH = – log [H3O+] = – log {4.2 × 10–3} = 3 – log 4.2  = 2.37

Now, let 1L of 1M ACOH solution be diluted to VL to double the pH and the conc. of diluted solution be C.

H3COOH + H2O H3COO– + H3O+

t = 0 C     0               0

– 1.8 × 10–5 1.8 × 10–5 1.8 × 10–5

——————————————————————

t=teq C– 1.8 × 10–5 1.8 × 10–5 1.8 × 10–5

New pH = 2 × old pH 

= 2 × 2.37 = 4.74

pH = – log [H3O+] = 4.74

∴ [H3O+] = 1.8 × 10–5

Ka =

1.8 × 10–5 =

∴ C = 3.6 × 10–5 L

on dilution 

M1V1 = M2V2

1M × 1L = 3.6 × 10–5 L × V2

∴ V2 = 2.78 × 104 L

Exercise 2 a) Solubility in moles per dm3

= = 2.0 × 10–4 

Since CaF2 Ca2+ + 2F–

Therefore, [Ca2+] = 2.0 × 10–4 

and [F–] = 2 × 2.0 × 10–4 

Hence, solubility product 

Ksp = [Ca2+] [F–]2 = (2.0 ×10–4) (4.0 × 10–4)2

= 32 × 10–12 

1. b) i) Solubility is in general expressed in gm/lt, 

so solubility of BaSO4 = 2.3 × 10–3 g /lt

Loss in weight of BaSO4 = amount of BaSO4 soluble 

⇒ %loss  = × 100 = 1.15%

1. ii) Now 0.01 NH2SO4 ≡01 ions 

≡ 0.005 M ions 

Now presence of prior to washing BaSO4 will suppress the solubility of BaSO4 (due to common ion effect). The suppression will be governed by Ksp value of BaSO4. Sof first calculate Ksp of BaSO4.

Solubility  of BaSO4 in fresh water  = 2 × 10–3 g/lt

≡ = Mol/lt = 9.85 × 10–6 M

Ksp = [Ba2+] = (9.85 × 10–6)2 = 9.71 × 10–11

Now let x be solubility in mol/lt

⇒ [Ba2+] in solution  = x mol/lt and in solution  

= (x + 0.005) mol/lt

Ionic product = [Ba2+] = (x) (x + 0.005)

Ksp = Ionic Product at equilibrium (saturation)

⇒ 9.71 × 10–11 = (x) 9x + 0.005) ⇒ x2 + 0.005 x  – 9.71 × 10–11 = 0 

⇒ x = = 1.94 × 10–8 mol/lt = 1.94 × 10–8 × 233.4 g/lt

⇒ 4.53 × 10–6 gm of BaSO4 are washed away

⇒ percentage  loss = = 2.26 × 10–3%

9. Solved Problems 

 

9.1 Subjective  

Problem 1: 20 mL of a weak monobasic acid requires 30 mL 0.2 M NaOH for the end point. The pH of solution upon addition of 12 mL 0.2 M NaOH during the titration was found to be 6.2. Find pH of the stock solution of the weak monobasic acid.

Solution: 30 mL 0.2 M NaOH ≡ 30 × 0.2 i.e 6 millimoles NaOH

Weak acid being monobasic and NaOH being monoacidic, they will react together in 1:1 molar proportion. Thus 20 mL solution of weak monobasic acid will contain 6 millimole of that acid.

Molarity (C) of weak acid solution = = 0.3

When 12 mL 0.2 M NaOH i.e 12 × 0.2 (= 2.4) millimole of NaOH is added to acid solution, 2.4 millimole of acid will react with 2.4 millimole of acid to form 2.4 millimole of sodium salt of the acid. The acid remaining unreacted will be 3.6 (= 6 – 2.4) millimole. The mixture will be an acid a buffer mixture, the pH of which will be as given below.

pH = pKa +

6.2 = pKa +

∴ pKa = 6.2 + 0.176 = 6.376

[H+] = αC = (neglecting α in comparison to unity)

pH = –log (Ka⋅C)1/2 = [pKa – logC] = [6.376 – log 0.3]

= [6.376 + 0.522] = 3.45

Problem 2: The pH 0.1 M aqueous solution of a 2° amine is 11 at 25°C. Calculate its Kb.

Solution: R2NH + H2O + OH–

Initial conc. 0.1 M Excess 0 0

Equilibrium conc. (0.1 – x)M Constant xM xM

pH + pOH = 14 (at 25°C)

pOH = 14 – 11 = 3

∴ [OH–] = 10–3M

Thus x = 10–3

The equilibrium constant of the above ionic equilibrium is base dissociation constant (Kb) of the 2°–amine given as

Kb = = 1.01 × 10–5

Problem 3: 100 mL 1.0 M HA (Ka = 2.0 × 10–5) and 100 mL 0.1 M HCl are mixed together. Find pH of the resulting mixture. What will be the effect of adding (a) 55.0 mL 1.0 M KOH solution is added and (b) 55.0 mL 2.0 M KOH solution, are added to the above acid mixture?

Solution: Upon mixing 100 mL each of 1.0 M HA and 0.1 M HCl, the conc. of each specie will be halved. HCl (a strong acid) will ionise completely and H+ ion thus produced will exert common ion effect on the dissociation equilibrium of HA, a weak acid as evident from its Ka.

HA H+ + A–

Initial conc. C 0 0

Equilibrium conc. C(1 – α) (Cα + 0.05) Cα

Ka =

α being very small due to common ion effect, may be neglected in comparison to unity. Thus

Ka = (Cα + 0.05)α

2.0 × 10–5 = (0.5α + 0.05)α

0.05 α2 + 0.05 α – 2.0 × 10–6= 0

α = 3.998 × 10–5

[H+] = Cα + 0.05

pH = 1.3 

55. a) pH upon addition of 55.0 mL 1.0 M KOH: 

Addition of 55.0 mL 1.0 M KOH means the addition of 55 millimole KOH. 10 millimole of this would first neutralise (100 mL 0.1 HCl) 10 millimole of HCl, present in the mixture. Remaining 45 millimole of will react with 45 millimoles of weak acid present in the mixture forming 45 millimoles of salt (KA). Total amount of weak acid present is 100 mL 1.0 M i.e. 100 millimole. Thus 55 millimole of weak acid will remain unreacted. The resulting mixture will be an acid buffer mixture with . The pH is, therefore, calculated as

pH = pKa + = 6 + = 5.91

ΔpH = 5.91 – 1.3 = 4.61

2. b) pH upon addition of 55 mL 2.0 M KOH: 

55 mL 2.0 M KOH i.e. 110 millimoles KOH will neutralize both the acids completely the total amount of which in the mixture is 110 millimoles. Also since both the acids are monobasic and KOH is monoacidic so reaction will occur in equimolecular proportion. Salt (NaA) formed will be 100 millimole.

The volume of solution = 255 mL

Molar conc. (C) of NaA in solution = = 0.392

The salt of weak acid with a strong base will hydrolyse to give free OH– in solution rendering the solution alkaline. The pOH of the solution will, therefore, be calculated using the formula of salt hydrolysis.

pOH = [pKw – pKa – logC]

= [14 – 6 – log 0.592]

= [14 – 6 + 0.406

= 4.203

pH = 14 – 4.203 = 9.79

ΔpH = 9.79 – 1.3 = 8.49

Problem 4: Determine the number of moles of AgI which may be dissolved in 1 L of 1 M CN– solution. Ksp for AgI and Kf for [Ag(CN)2]– are 1.2 × 10–7 M2 and
7.1 × 1019 M–2 respectively.

Solution: The relevant equilibrium with problem and the overall equilibrium is as given below.

AgI(s)  Ag+ + I– Ksp = 1.2 × 10–7 M2

Ag+ + 2CN– [Ag(CN)2]– Kf = 7.1 × 1019 M–2

––––––––––––––––––––––––––––––––––––––––––––

AgI(s) + 2CN– [Ag(CN)2]– + I– K = 1.2 × 7.1 × 1012

Initilaly 1 M 0   0   = 8.52 × 1012

At equilibrium (1 – 2x)M xM xM

Where x is the solubility of AgI in mole L–1

= 8.52 × 1012

⇒ = 2.91 × 106

⇒ x = 2.91 × 106 – 5.82 × 106x

⇒ x + 5.82 × 106x = 2.91 × 106

⇒ 5.82 × 106x = 2.91 × 106

⇒ 5.82 × 106x = 2.91 × 106  ⇒ x = = 0.5

Another method: 

Let the solubility of AgI be x moles/ lit 

AgI(s)  Ag+(aq) + I–(aq)

x x

Then Ag+ combines with CN– to produce Ag(CN)–2. Since Kf for the complex is high, we can assume that almost all of Ag+ combines 

Ag+ + 2CN–  

1–2 x x

Kf =

Moreover 

[Ag+]  [I–] = KSP

⇒[Ag+] x = KSP

⇒Kf = =

∴ x = 0.49 moles 

Problem 5: A biochemical experiment is to be carried in NH4OH – NH4Cl buffer mixture of pH = 9.26. What volume of 2.0 M NH4Cl solution needs to be added to 20 mL 5.0 M NH4OH solution so as to maintain the pH at the above value.

Kb for NH4OH = 1.8 × 10–5

Solution: The mixture is a base buffer mixture for which the formula to be applied is

pOH = pKb +

pOH = 14 – pH = 14 – 9.26 = 4.74

pKb = – log (1.8 × 10–5) = 4.74

Let V ml of  2M NH4Cl solution needs to be added.

∴

Using the above formula

4.74 = 4.74 + ⇒ = 0  ⇒ = 1 ⇒ v = 50

Thus 50 mL of 2 M NH4Cl will have to be added.

Problem 6: When 0.01 M solution of MnCl2 that has been adjusted to pH 3.0 with HCl, is saturated with H2S (0.10 M), should MnS precipitate?

For H2S: Ka1 = 1.1 × 10–7

Ka2 = 1.00 × 10–14

Ksp(MnS) = 1.4 × 10–15  

Solution: Ka(H2S) = Ka1× Ka2 = = 1.1 × 10–7 × 1.0 × 10–14 = 1.1 × 10–21

Now, at pH = 3; [H+] = 10–3 M and small contribution of H+ ions from H2S can be neglected.

∴ = 1.1 × 10–21

or [S2–] = 1.1 × 10–16

Now, Ksp = [Mn2+] [S2–] = 1.4 × 10–15

And ionic product = (0.01) (1.1 × 10–16) 

= 1.1 × 10–18 [0.01 (M) Mn2+ coming from 0.01 (M) Mn(NO3)2]

Since ionic product does not exceed the Ksp, no precipitation occurs

Problem 7: K1 and K2 of carbonic acid are 4.3 × 10–7 and 5.6 × 10–11 respectively. Calculate the pH of 0.1 M Na2CO3 solution.

Solution: In solution Na2CO3 will be completely ionised and CO32– ion, a strong base for being conjugate of very weak acid HCO3–, will hydrolyse to give free OH– ion in solution.

1. CO32– + H2O HCO3– + OH–, = 1.8 ×10–4

HCO3–, is also a strong base for being conjugate base of a weak acid, H2CO3, will also hydrolyse to give free OH– ion in solution.

2. HCO3– + H2O H2CO3 + OH–, Kb= = 2.8 × 10–8

 

Where are the hydrolysis constants of CO32– and HCO3– respectively.

Since , the second step hydrolysis can be neglected.

So considering only the 1st step hydrolysis equilibrium we have

CO32– + H2O HCO3– + OH—

C excess 0 0

C(1 – h) constant Ch Ch

= ≈ h2C (∴ h << 1)

h =

[OH–] = Ch =

= = 4.24 × 10–3M

pOH = – log (4.24 × 10–3) = 2.37

∴ pOH = 14 – 2.37 = 11.62

Alternatively we may use the formula directly:

pOH = [pKw – pK2 – logC]

= [14 – 10.25 – log0.1]

= [14 – 10.25 + 1] = 2.37

∴ pH = 11.62

‘Problem 8: The solubility product of Ag2C2O4 at 25°C is 1.29 × 10–11 mole3L–3. A solution of K2C2O4 containing 0.152 moles in 500 mL water is shaken at 25°C with excess Ag2CO3 till the following equilibrium is reached.

Ag2CO3(s) + K2C2O4(aq) Ag2C2O4(s)↓ + K2CO3(aq)

At equilibrium, the solutions contains 0.0358 mole of K2CO3. Assuming the degree of dissociation of K2C2O4 and K2CO3 to be equal, calculate the solubility product of Ag2CO3.

Solution: Initial conc. of K2C2O4 solution = 0.152 × 2 = 0.304 M

Conc. of K2CO3 at equilibrium = 0.0358 × 2 = 0.0716 M

The equilibrium given in the problem is the overall of the two equilibria as shown below:

Ag2CO3(s)  2Ag+ + CO32– Ksp = ?

2Ag+ + C2O42– Ag2C2O4(s)

-––––––––––––––––––––––––––––––––––––––––––––––––––

Ag2CO3(s) + C2O42– Ag2C2O4(s) + CO32– K =

Initial conc. 0.304 M 0

(0.304 – 0.0716)M 0.0716 M

(Assuming complete dissociation of K2C2O4 and K2CO3 both)

= = = 0.308

∴ Ksp(Ag2CO3) = 0.308 × 1.29 × 10–11 = 3.97 × 10–12 mole3 L–3

Problem 9: How many mole of HCl will be required to prepare one litre of a buffer solution containing HCN and NaCN of pH 8.5 using 0.01 mole of NaCN?

Ka(HCN) = 4.1 × 10–10

Solution: pH = pKa+

Let x mole of HCl be added. x mole HCl will react with x mole NaCN to form x mole HCN.

NaCN + HCl ⎯→ NaCl + HCN

More correctly,

CN– + H+ ⎯→ HCN

0.01 x 0

0.01 – x 0 0

Thus,

8.5 = – log (4.1 × 10–10) +

⇒  = – 0.8872

⇒ = 0.1296

⇒ x = 8.85 × 10–3M

Problem 10: The Ksp for Ca(OH)2 at 25°C is 4.42 × 10–5. A 500 mL of saturated solution of Ca(OH)2 is mixed with equal volume of 0.4 M NaOH. How much Ca(OH)2 in milligrams is precipitated?

Solution: Let the solubility of Ca(OH)2 is pure water be S mole L–1

Ca(OH)2(s) Ca2+ + 2OH–

s 2s

Ksp= [Ca2+] [OH–]2 

4.42 × 10–5 = s(2s)2 = 4s3

∴ s = = 0.0223 mole L–1 

No. of mole of Ca2+ ion in 500 mL of saturated solution =  

= 0.01115

Upon addition of 500 mL of 0.4 M NaOH to 500 mL of saturated solution of Ca(OH)2, the concentration of NaOH i.e. OH– ion will be halved. That is

[OH–] = = 0.2 M

[Ca2+] = = 0.001105 M

Thus, no. of mole of Ca2+ or Ca(OH)2 ppted out

= 0.0115 – 0.001105 = 0.010045

Mass of Ca(OH)2 ppted = 0.010045 ×74 

  = 0.7433g = 743.3 mg

Problem11: Calculate the pH at the equivalence point of the titration between 0.1M
CH3COOH (25 ml) with 0.05 M NaOH. Ka (CH3COOH) = 1.8 × 10–5.

Solution: We have already seen that even though when CH3COOH is titrated with NaOH the reaction does not go to completion but instead  reaches equilibrium. We can assume that the reaction is complete and then salt gets hydrolysed because, this assumption will help us to do the problem easily and it does not effect our answer.

∴ [H+] =  

First of all we would calculate the concentration of the salt, CH3COONa.  For reaching equivalence point, 

N1V1 = N2V2

∴ 0.1 × 25 = 0.05 × V2  ⇒ V2 = 50 ml 

Therefore [CH3COONa] =

∴[H+] = = 2.32 × 10–5

⇒ pH = – log 2.32 × 10–5 = 8.63

Problem 12: Find the concentration of H+, and , in  a0.01M solution of carbonic acid if the pH of this is 4.18

Ka1(H2CO3) = 4.45 × 10–7 and Ka2 = 4.69 × 10–11

Solution: pH = – log[H+]

4.18 = – log [H+]

[H+] = 6.61 × 10–5

H2CO3 H+ +

Ka =

or, 4.45 × 10–7 =

∴ = 6.73 × 10–5

again, H+ +

∴ Ka2 = = 4.69 × 10–11 =

∴ = 4.8 × 10–11

Problem 13: What is solubility of PbS (a) ignoring the hydrolysis of ions (b) including the hydrolysis of ions (assume pH of solution  = 7). 

Given that:

Ksp(pbs) = 1 × 10–7, Ka2(H2S) = 10–14

Solution: a) Pbs(S) Pb++ + S––

Ksp = [Pb++] [S––]

= S × S = S2 = 7 × 10–29

∴S = 8.4 × 10–15

1. b) Including hydrolysis: The equilibria of interest are 

The equilibria of interest are 

Pb2+ + 2H2O Pb(OH)+ + H3O+ ; Kh1 =  

= —- (i)

= = 6.7 × 10–7

S– – + H2O HS– + OH– , = 1  (ii)

HS– – + H2O HS– + –OH; = = 9.1 × 10–8 

= —– (iii)

Mass balance expression are:

[Pb2+]o = [Pb2+] + [Pb(OH)]+ …(a)

[S—]o = [S––] + [HS–] + [H2S] …(b)

Substituting the value of [Pb(OH)+] from equation (i) into equation (a)

[Pb++]o = [Pb++] + = [Pb++] =

∴[Pb++] = …(c)

Similarly using equation (ii) and (iii) in equation (b) we have 

[S– –] = …(d)

Now, PbS(s) Pb++ + S—

Ksp = [Pb++] [S––]

Substituting the values of [Pb++] and [S––] from equations (c) and (d), we get 

Ksp(Pbs) =

If y be the amount of PbS dissolve then 

[Pb2+]o = [S––]o =  [pH = 7, ∴ [H3O+] = 10–7 = [–OH]]

∴Ksp(PbS) = = 7 × 10–29

On solving, Y = 1.0146 × 10–10

 

Problem 14: Will Fe(OH)3(s) precipitate from a buffer solution prepared by mixing 0.5M CH3COOH and 0.15M CH3COONa at 25°C, if solution contains 0.25M Fe3+? KspFe(OH)3  = 4 × 10–38, Ka(CH3COOH) = 1.74 × 10–5.

Solutions: We have,

[H+] = Ka = 1.74 × 10–5 × -= 5.8 × 10–5

∴ [OH–} = + = 1.7 × 10–10

Now, ionic product of Fe(OH)3 = [Fe3+] × [OH–]3 = 0.25 × [1.7 × 10–10)3

 Q > Ksp, the precipitate will form 

Problem 15: Assuming the complete dissociation of HCl and the lead salt, calculate how much HCl is added to 0.001M lead salt solution to just percent precipitation when saturated with H2S. The concentration of H2S in its saturated solution is 0.1M

Ka (H2S) = 1.1 × 10–23

Ksp (PbS) = 3.4 × 10–28

Solution: We know,

Ksp(PbS) = [Pb+2] [S–2]

Since lead salt is completely dissociated, [Pb+2] is equal to the concentration of lead salt, i.e. [Pb+2] = 0.001M. If[S–2] is the concentration of S–2 required to just start precipitation of PbS.

[S–2] = = 3.4 × 10–25

Now the addition of HCl with suppress the dissociation of H2S to that extent that [S–2] = 34 × 10–25 (M)

 HCl is completely ionised, ∴ [H+] = [HCl]

Let [HCl] be x′. Therefore [H+] = x′

H2S 2H+ + S–2

At equilibrium [H2S] = 0.1 – 3.4 × 10–25 ≈ 0.1

“ [H+] = 2 × 3.4 × 10–25 + x′ ≈ x′

[S–2] = 3.4 × 10–25

Ka =

1.1 × 10–23 =

∴ x′ = 1.80

This any concentration of HCl greater than 1.8 M

Will just prevent precipitation

9.2 Objective 

Problem 1: An acidic indicator HIn (Kin = 10–6) ionises as HIn H+ + In–. The acid colour predominates over the basic colour when HIn is at least 10 times more concentrated than In– ion. On the other hand basic colour predominates over the acid colour when the In– ion is at least 5 times more concentrated than HIn. Hence pH range of the indicator is

(A) 5.0 – 6.7 (B) 7.0 – 8.7

(C) 5.3 – 7.0 (D) 7.0 – 8.1

Solution: pH = pKin + , where pKin = 6

For acid colour to predominates

∴ pH ≤ 6 – 1 i.e 5.0

for base colour to predominate ≥ 5

∴ pH ≥ 6 + log5 ⇒ pH ≥ 6.7

∴ (A)

Problem 2: The correct statement amongst the following is

(A) A strong electrolyte remains completely dissociated at all dilutions

(B) Upon dilution the degree of dissociation of a weak electrolyte and number of ions per unit volume of its solution both increase.

(C) A strong electrolyte is completely ionised at all dilutions but not completely dissociated.

(D) pH of solution of a weak acid decreases with dilution.

Solution: “Complete dissociation” implies that interionic attraction has completely ceased to exist. This condition in the case of a solution of strong electrolyte is achieved only at infinite dilution when concentration of solution tends to zero.

∴ (C)

Problem 3: 4M solution of a weak monobasic acid (x% ionized and pH = 3.0) is diluted to 1 M by adding water (distilled). Percentage ionisation and pH of solution after dilution will be respectively.

(A) 2x and 2.7 (B) 0.25x and 3.3

(C) 0.5x and 2.7 (D) 2x and 3.3

Solution: Solution of weak acid being concentrated, we can use the approximate

α = i.e. α ∝

Thus, decreasing the conc. to one fourth, α will be doubled. Doubling of α means doubling of percentage ionisation. Hence % ionisation will be 2x.

[H+] = i.e. [H+] ∝

If the conc. is decreased to one fourth of its original value H+ ion conc. will be halved. Thus, after dilution [H+] = 0.5 × 10–3M

∴ pH = 3.3

∴ (D)

Problem 4: pH of a buffer solution changes from 6.20 to 6.17 when 0.003 mole of acid is added to 500 mL of the buffer. The buffer capacity of the system is, therefore 

(A) 0.1 (B) 0.3

(C) 0.2 (D) 0.4

Solution: Buffer capacity =

= = 0.2

∴ (C)

Problem 5: Ksp of CaSO4 is 2.4 × 10–5 at 25°C. In a solution containing Ca2+ ions the precipitation of CaSO4 begins to occur when SO42– ion concentration in the solution is made just to exceed the value of 4.8 × 10–3M. Hence concentration of Ca2+ ion in the solution  is

(A) 200 ppm (B) 40 ppm

(C) 400 ppm (D) 100 ppm

Solution: [Ca2+] [SO42–] = 2.4 × 10–5

[Ca2+] = = 5 × 10–3M

1 L solution contains 5 × 10–3 mole of Ca2+ ions

1000 L solution will contain 5 mole i.e. 200 g Ca2+ ions

Taking density of aqueous solution to be unity

1000 kg i.e. 106 g solution contains 200g Ca2+ ions

Conc. of Ca2+ ion = 200 ppm

∴ (A)

Problem 6: Correct statement regarding pure water amongst the following is

(A) It contains only single specie i.e. H2O molecules

(B) It contains three species: H2O (molecules), H+ and OH–

(C) It contains only two species H3O+ and OH–

(D) It contains three species H2O(molecules), H3O+ and OH–

Solution: Water remains feebly ionised into H+ and OH– ions. H+ ion does not exist independently but as H3O+ as shown below:

H2O + H2O H3O+ + OH–

∴ (D)

Problem 7: There is a solution which is one molar w.r.t. each M2+ and X3+ ions present in it. The Ksp of M(OH)2 and X(OH)3 are 4.0 × 10–10 and  2.7 × 10–14 respectively . If NH4OH solution is added gradually to the above solution which of the following will happen?

(A) Both M(OH)2 and X(OH)3 will precipitate together

(B) M(OH)2 will precipitate first.

(C) X(OH)3 will precipitate first

(D) None of these will precipitate with NH4OH solution

Solution: M(OH)2 will precipitate when

[M2+] [OH–]2 > 4.0 × 10–10

[OH–] > 2.0 × 10–5 molar ( [M2+] = 1)

X(OH)3 will precipitate when

[X3+] [OH–]3 > 2.7 × 10–14

∴ [OH–]  > 3 × 10–5 molar ( [X3+] = 1)

Thus M(OH)2 requires less concentration of OH– ion for precipitation. Hence M(OH)2 will precipitate first.

∴ (B)

Problem 8: 20 mL of a weak acid HX is titrated against 0.1 M NaOH. At the point of half equivalence the pH of solution is 5.7. Hence Ka of acid is

(A) 7.0 × 10–5 (B) 2.0 × 10–5

(C) 7.0 × 10–4 (D) 2.0 × 10–6

Solution: At the point of half neutralization the mixture will be an acid buffer with = 1

pH = pKa + log1

pKa = 5.7

Ka = 2.0 × 10–6

∴ (D)

Problem 9: If an aqueous solution at 25°C has twice as many OH– as pure water its pOH will be

(A) 6.699 (B) 7.307

(C) 7 (D) 6.98

Solution: [OH–] = 2 × 10–7

pOH = 14 – pH or – log [OH–] 

∴(A)

Problem 10: Solubility of AgCl in water, 0.01M CaCl2, 0.01M NaCl and 0.05M AgNO3 are S1, S2, S3 and S4 respectively then.

(A) S1 > S2 > S3 > S4 (B) S1 > S3 > S2 > S4

(C) S1 > S2 = S3 > S4 (D) S1 > S3 > S4 < S2

Solution: AgCl Ag+ + Cl–

In CaCl2 

CaCl2 Ca+2 + 2Cl–

0.01 0.01 2×0.01

In NaCl

NaCl Na+ + Cl–

0.01 0.01 0.01

In AgNO3

AgNO3 Ag+ + Cl–

0.05 0.05 0.05

common ion effect is maximum in AgNO3

So, S1 > S3 > S2 > S4

∴(B)

Problem 11: pH of Ba(OH)2 solution is 12. Its solubility product is 

(A) 10–6M3 (B) 4 × 10–6M3
(C) 0.5 × 10–7M3 (D) 5 × 10–7M3

Solution: Since pH = 12 ∴pOH = 14 – 12 = 2

∴[OH–] = 10–2M

We know Ba (OH)2 Ba++ + 2OH–

∴[Ba++] = M

∴KSP = [Ba++] [OH–]2 = × (10–2)2 = 5 × 10–7M3

∴(D)

Problem 12: he hydrolysis constant for ZnCl2 will be 

(A) Kh = (B)

(C) Kh = (D)

Where Kb is effective dissociation constant of base Zn++

Solution: Zn++ + 2H2O  Zn(OH)2 + 2H+

∴ Kh = … (1)

Zn(OH)2     Zn++ + 2OH–

∴Kb = , Kw = [H+] [OH–]

∴ = Kh 

∴ (B)

Problem 13: In which case pH will not change on dilution 

(A) 0.01M CH3COONa + 0.01M CH3COOH buffer 

(B) 0.01M CH3COONH4

(C) 0.01M NaH2PO4

(D) in all cases 

Solution: Mixture of sodium acetate and acetic acid is a buffer of pH value equal to pKa so its buffer capacity is very high  and hence its pH will not change significantly while CH3COONH4 is a salt of weak acid CH3COOH and weak base NH4OH whose magnitude of Ka and Kb are equal. So its pH does not depend upon concentration. Further more, NaH2PO4 is, in fact, a single solute buffer

∴(D) 

Problem 14: M(OH)x has KSP 4 × 10–12 and solubility 10–4M. Then the value of x is 

(A) 1 (B) 2

(C) 3 (D) –4

Solution: M(OH)x will ionize in the way 

M(OH)x M+ x + x OH–

  10–4 x × 10–4

∴ Ksp = [M+x] [OH–]x

∴ (10–4) (x × 10–4)x = 4 × 10–12

by inspection we get this relation will hold good when x = 2

∴(B)

Problem 15: pH of a mixture of 1M benzoic acid (pKa = 4.20) and 1M C6H5COONa is 4.5. In 300 ml buffer, benzoic acid is [log 2 = 0.3]

(A) 200 ml (B) 150 ml 

(C) 100 ml (D) 50 ml 

Solution: pH = pKa + log

∴ 4.5 = 4.2 + log   = log = 0.3 

(since log 2 = 0.3)

∴ = 2 

Let V ml 1M C6H5 COOH solution and  (300 – V) 1M C6H5COONa  solution be mixed together

[Acid] = ;[Salt] =

[Acid] = ; [Salt] =

∴ = 2 

= 300 – V = 2V ∴ V = 100 ml

∴(C)

 

10. Assignments (Subjective Problems)

 

LEVEL – I

1. The pH of 0.1 M solution of a weak monoprotic acid is 4.0. Calculate pH of the solution obtained by diluting 10 mL of this solution with 990 mL water.

 

2. The self ionization constant of pure formic acid, K = [HCOOH2+] [HCOO–] has been estimated as 10–6 at room temperature. What percentage of formic acid molecules in pure formic acid are converted to formate ion? The density of formic acid is 1.22g/cm3.
3. Calculate the change in pH resulting from the addition of (a) 0.02 moles of dissolved gaseous HCl and (b) 0.02 mole of dissolved NaOH to 1 litre buffer solution containing 0.1 moles each of NH3 and NH4Cl.

Kb(NH3) = 1.8 × 10–5

4. How many grams of NH4Cl should be dissolved in 250 ml of water to have a solution of pH 5.0? Kb for NH4OH is 1.8 × 10–5.
5. For 0.50 M aqueous solution of sodium cyanide, (pKb of CN– of 4.70), calculate (i) hydrolysis constant, (ii) percentage hydrolysis and (iii) pH.
6. The solubility of Pb(OH)2 in water is 6.7 × 10–6M. Calculate the solubility of Pb(OH)2 in a buffer solution of pH 8.
7. Calculate amount of ammonium sulphate in gram which must be added to 500 mL of 0.2 M NH3 to yield a solution of pH = 9.35. Kb for NH3 = 1.78 × 10–5.
8. A solution is saturated with respect to SrCO3 and SrF2. The was found to be 1.2 × 10–3 mol L–1. Determine F– ion concentration. Given : Ksp (SrCO3)  = 7.0 × 10–10 and Ksp(SrF2) = 7.9 × 10–10.

9. Bromophenol blue is an indicator with a Ka value of 5.84 × 10-5. What % of this indicator is in its basic form at a pH of 4.84 ?

10. 11. 4 g of Potassium alum is dissolved in enough water to make 100 ml solution. Calculate its [H+] ion concentration.

Al3+ + 2H2O Al(OH)++ + H+  Kh = 14 × 10–5

 

LEVEL – II

1. Calculate the amount of NH3 and NH4Cl required to prepare a buffer solution of pH 9.0, when total concentration of buffering reagents is 0.6 mol L–1. What will be the effect of adding 2g NaOH in 500 mL of this buffer, on pH ?

pKb for NH3 = 4.7

2. 20 mL 1.0 M solution (pH = 3) of a weak monobasic acid is titrated with 0.5 N solution of NaOH. Calculate pH of the solution when (a) 10 mL, (b) 20 mL and (c) 40 mL of NaOH solution have been added.
3. Two buffers (X) and (Y) of pH 4.0 and 6.0 respectively are prepared from acid HA and the salt NaA. Both the buffers are 0.50 M in HA. What would be the pH of the solution obtained by mixing equal volumes of two buffers?

Ka of HA = 1.0 × 10–5.

4. Determine the concentration of NH3 solution whose one litre can dissolve 0.10 mole of AgCl. Ksp of AgCl and Kf of Ag(NH3)2+ are 10–10 M2and 1.6 × 107 M–2 respectively.
5. Calculate the pH of 0.1 M K3PO4 solution. K1, K2 and K3 of H3PO4 are 5.0 × 10–3,
   2.5 × 10–8 and K3 = 1.0 × 10–12.
6. 25 mL of a weak base BOH was titrated with 0.5 M HCl. The pH of solution upon addition of 10 mL acid was 8.6 and that upon addition of 25 mL acid was 8. Calculate pH of the base solution when 0 mL acid has been added. Also calculate the pH at the end point.
7. The Ksp of Mg(OH)2 is 8.9 × 10–12 at 25°C. If the pH of solution is adjusted to 9.0. How much Mg2+ ion will be precipitated as Mg(OH)2 from a 0.1M MgCl2 solution at 25°C? Assume that MgCl2 is completely dissociated.
8. An aqueous solution contains 10% ammonia by mass and has density of
   0.99 gm cm–3. Calculate hydroxyl and hydrogen ion concentration in this solution. Ka for = 5 × 10–10 M.
9. A solution of 0.010 M CdCl2 contians 0.010 (M) NH3. What conc. of +NH4 ion from NH4Cl is necessary to prevent the precipitation of Cd(OH)2? (Ksp = 2 × 10–14 and Kb = 1.8 × 10–5)
10. An aqueous solution of aniline of conc. 0.24 M is prepared. What concentration of sodium hydroxide is needed in this solution, so that anilium ion conc. remains at 1 × 10–8 M. Ka = 2.4 × 10–5.

 

LEVEL – III

1. 12.5 mL solution of 0.5 M H3XO4 is titrated with 0.5 M NaOH. Calculate pH at the following stages.
2. a) When 1st step neutralization is complete
3. b) When 2nd step neutralization is complete
4. c) When 3rd step neutralization is complete

K1, K2 and K3 for H3XO4 are 1 × 10–4, 2.0 × 10–7 and 4.0 × 10–11 respectively.

2. 25 mL of a dilute aqueous solution p-hydroxybenzoic acid is titrated with 0.02 M NaOH solution. The solution has pH = 4.57 when 8.12 mL of alkali had been added, and pH = 7.02 after 16.24 mL (the equivalent point) had been added. Use these data to find out Ka1 and Ka2 for p-hydroxybenzoic acid.
3. An acid base indicator which is actually a weak monobasic acid exists 50% dissociated in a solution of HCl which is 4.0 × 10–5M. The acid colour and basic colour of the indicator are distinct and say they are colour A and colour B respectively. Colour A predominates over colour B when concentration of species possessing it is 8 times that of the species possessing colour B. On the other hand colour B predominates over colour A when concentration of the species possessing it is 20 times that 4 the species possessing colour A. Calculate pH range of indicator.
4. A weak base BOH was titrated against a strong acid. The pH at 1/4th equivalence point was 2.94. Enough strong base was now added (6 milli equivalent) to completely convert the salt. The taol volume was 50 mL. Find pH at this point.
5. The ionisation constant of NH4+ ion in water 5.6 × 10–10 at 25°C. The rate constant for the reaction of NH4+ + O–H to form NH3 and H2O at 25°C is 3.4 × 1010 lit/mol sec. Calculate the rate constant for proton transfer from water to NH3
6. To 100 ml of a solution, which contains 8.29 × 10–3 lead ions, 100 ml of 10–3 M H2SO4 is added. How much lead remains in the solution unprecipitated? (Ksp of PbSO4 = 2.2 × 10–8).
7. Calculate the molar solubility of Zn(OH)2 in 1 (M) NH3 solution at room temperature

Ksp of Zn(OH)2 = 1.8 ×  10–17

K stability [Zn (NH3)4]2+= 1.64 × 1010

8. The solubility of Mg(OH)2 is increased by addition of NH4+ ion. Calculate
9. i) Kc for the reaction

Mg(OH)2 + 2NH4+ 2NH3 + 2H2O + Mg2+

1. ii) Find solubility of Mg(OH)2 in a solution containing 0.5 M NH4Cl

(Ksp Mg(OH)2) = 1.0 × 10–11, Kb(NH3) = 1.8 × 10–5)

9. A solution contains a mixture of Ag+(0.1M) and (0.10M) which are separate by selective precipitation. Calculate maximum concentration of  Iodide ion at which one of them gets precipitated almost completely. What percentage of that metal ion is precipitated? [Ksp of AgI = 8.5 × 10–17, Ksp of Hg2I = 2.5 × 10–26]
10. Calculate (derived from NH4Cl) needed to prevent Mg(OH)2 from precipitating in a litre of solution which contains 0.02 mole of ammonia and 0.001 mole of Mg+ ions. The ionisation constant of ammonia is 1.8 × 10–5 and Ksp of Mg(OH)2
    is 1.12 × 10–11.
11. 1.0 L of solution which was in equilibrium  with solid mixture of AgCl and Ag2CrO4 was found to contain 1 × 10–4 moles of Ag+ ions, 1.0 × 10–6 moles of Cl– ions and 8.0 × 10–4 moles of ions. Ag+ ions are added slowly to the above mixture (keeping volume constant) till 8.0 × 10–7 moles of AgCl got precipitated. How many  moles of Ag2CrO4 were also precipitated?
12. HN3 (hydrazoic acid) is a weak acid dissociating as: HN3 ⎯⎯→ H+ + . Find the concentration of Ag+ ions, if excess of solid AgN3 is added to HN3 at pH = 4. The ionisation constant of HN3 is 1.9 × 10–5. The solubility of AgN3 in pure water is found to be 5.4 × 10–3 M at 25°C.
13. Calculate approximate pH of a 0.1 M aqueous solution of H2S, K1 and K2 for H2S are 1.00 × 10–7 and 1.3 × 10–3 respectively at 25°C.
14. Calculate sulphide ion concentration in saturated H2S solution in water at 25°C when the pH of the solution is 11. The molarity of saturated H2S solution in water is 0.1M and K1, K2 for H2S are 5.7 × 10–8 and 1.2 × 10–15 at 25°C.
15. Two weak monobasic organic acids HA and HB have dissociation constants as 1.5 × 10–5 and 1.8 × 10–5 respectively at 25°C. If 500 ml of 1 M solutions of each of these two acids are mixed to produce 1 litre of mixed solution, what is the pH of the resulting solution?

 

11. Assignments (Objective Problems)

 

LEVEL – I

1. If the degree of ionization of water be 1.8 × 10-9 at 298K. Its ionization
   constant will be

(A) 1.8 × 10-16 (B) 1 ×10-14

(C) 1 × 10-16 (D) 1.67 × 10-14

2. When a solution of benzoic acid was titrated with NaOH the pH of the solution when half the acid neutralized was 4.2. Dissociation constant  of the acid is

(A) 6.31 × 10-5 (B) 3.2 × 10–5

(C) 8.7 × 10–8 (D) 6.42 × 10–4

3. 10-2 mole of NaOH was  added to 10 litre of water. The  pH  will change by

(A) 4 (B) 3

(C) 11 (D) 7

4. For an aqueous solution to be neutral it must have

(A) pH = 7 (B) [H+]=[OH–]

(C) [H+] = (D) [H+] < [OH–] 

5. If an aqueous solution at 25°C has twice as many OH– as pure water its pOH will be

(A) 6.699 (B) 7.307

(C) 7 (D) 6.98

6. What would be the pH of an ammonia solution if that of an acetic acid solution of equal strength is 3.2? Assume dissociation constant for NH3 & acetic acid are equal.

(A) 3.2 (B) 6.4                  

(C) 9.6                 (D) 10.8

7. The pH of an aqueous solution of 0.1M solution of a weak monoprotic acid which is 1% ionised is 

(A) 1 (B) 2

(C) 3 (D) 11

8. pH of a 10–10 M NaOH is nearest  to 

(A) 10 (B) 7

(C) 4 D) 10

9. The weight CH3COONa needed to add in 1 litre CH3COOH to obtain a buffer solution of pH value 4 is (Given Ka = 1.8 × 10–5)

(A) 1.2 g (B) 1.47g

(C) 1.8 g (D) 4 g 

10. A dilute HCl solution saturated with H2S has pH value 3 then  the conc. of S– – is (Given K1 = 1 × 10–7,  K2 = 1.3 × 10–13)

(A) 2 × 10–13 (B) 2.4 × 10–13

(C) 3 × 10–15 (D) 1.3 × 10–15 M

11. Which is the strongest acid

(A) H3AsO4 (B) H2AsO4–

(C) HAsO4— (D) AsO4—

12. In an aqueous solution of triprotic acid H3A which is true

(A) [H+] = 3[A3–] (B) [H+] > 3[A3–]

(C) [H+] < 3[A3–] (D) [H+] = [OH–]

13. If K1 & K2 be first and second ionisation constant of H3PO4 and K1>> K2 which is incorrect.

(A) [H+] = [H2 PO4–] (B) [H+] =  

(C) K2 = [HPO4– –] (D) [H+]= 3[A3–]

14. Which one of the following anion does not hydrolyse

(A) H– (B) CN–

(C) NO2– – (D) S– –

15. If the ionic product of water varies with temperature as follows and the density of water be nearly constant for this range of temperature the process
    H+ + OH H2O is

Temp. °C	0	10	25	40	50
Kw	0.114×10–14	0.292×10–14	1.008 ×10–14	2.91×10–14	5.474 ×10–14

(A) Exothermic (B) Endothermic

(C) Can’t say (D) Ionization

 

LEVEL – II

1. 0.1M solution of which of the following substances is most acidic

(A) NH4Cl (B) KCN

(C) AlCl3 (D) NaC2H3O2

2. The following reaction takes place in the body

CO2 + H2O H2CO3 H+ + HCO3–. If CO2 escapes from the system

(A) pH decreases (B) [H+] will decrease

(C) [H2CO3] remains the same (D)forward reaction will be promoted

3. The solubility of A2X5 is x mole dm-3. Its solubility product is

(A) 36x6 (B) 64 x 104x7

(C) 126 x 7 (D)1.25 x 104x7

4. The solubility of CH3CO2Ag would be least amongst the following
   solvents in

(A) acidic solution of pH = 3 (B) basic solution of pH = 8

(C) neutral  solution of pH = 7 (D) pure water

5. If Kh (hydrolysis constant) for anilinium ion is 2.4 x 10-5M, then Kb for
   aniline will be

(A) 4.1 x 1010 (B) 4.1 x 1010

(C) 2.4 x 109 (D) 2.4 x 10-19

6. Acetic acid dissolved in ammonia will be

(A) Highly conducting (B) Less conducting

(C) Can’t say (D) More conducting than that in water

7. The equilibrium constant of the reaction:

A– + H2O HA + OH–

is 108 times more than ionic product of water at 25°C. Hence Ka of weak acid will be

(A) 10–8 (B) 10–6

(C) 10–14 (D) None of these

8. If we plot α2 verses volume V we will get 

(A) a straight line with slope value equal to Kdissociation  

(B) a  straight line with slope equal to Keq

(C) an exponential curve

(D) a parabola 

9. Equal volumes of two solutions of a strong acid having pH 3 and pH 4 are mixed together. The pH of the resulting solution will then be equal to

(A) 3.5 (B) 3.26

(C) 7 (D) 1.0

10. When equal volumes of following solution are mixed, precipitation of AgCl
    (Ksp = 1.8 x 10-10) will occur only with

(A) 10-4M; Ag+ and 10-4M; Cl– (B)10-5M; Ag+ and 10-5M; Cl–

(C) 10-6M; Ag+ and 10-6M; Cl– (D)10-10M; Ag+ and 10-10M; Cl–

11. Let Kw at 100°C be 5.5 × 10-13 M2. If an aqueous solution at this temperature has
    pH = 6.2. Its nature will be

(A) acidic (B) alkaline

(C) neutral (D) can’t say

12. Dissociation constant of two acids HA & HB are respectively 4 × 10-10 & 1.8 × 10-5 whose pH value will be higher for a given molarity:

(A) HA (B) HB

(C) Both same (D) Can’t say

13. pH value of pure water at 0°C will be

(A) Greater than 7 (C) Less than 7

(C) 7 (D) All the three

14. The pH of a buffer is 6.745. When 0.01 mole of NaOH is added to 1 litre of it, the pH changes to 6.832. Its buffer capacity is

(A) 0.187 (B) 0.0115

(C) 0.076 (D) 0.896

15. The aqueous solution  of potash alum is acidic due to hydrolysis of

(A) K+ (B) Al3+

(C) SO4– –- (D) presence of acid in its crystal as impurity 

 

12. Answers to Subjective Assignments

 

LEVEL – I

1. pH = 5 2. 0.004%
2. a) ΔpH = – 0.18; b) ΔpH = 0.1761 4. 2.43gm
3. Kh = 2 × 10–5, % Hydrolysis = 0.63, pH = 11.5
4. 1.2 × 10–3 M 7. 5.26g
5. 3.68 × 10–2M 9. 80%
6. 1.82 × 10–3 M

LEVEL – II

1. [NH4Cl] = 0.4 mole L–1, [NH3] = 0.2 mole L–1, ΔpH = 0.3
2. a) pH = 5.523, b) pH = 6, c) pH = 9.77
3. pH = 5.70 4. [NH3] = 2.7 M
4. pH = 12.5 6. pH = 11, pH = 4.239
5. 0.011 mole L–1 8. [OH–] = 0.0179, [H+] = 9.268 × 10–12M
6. 0.13 M 10. 10–2M

LEVEL – III

1. a) pH = 5.35, b) pH = 8.55, c) PH = 11.5
2. Ka1= 2.7 × 10–5, Ka2= 3.4 × 10–10 3. pH range = 3.5 – 5.7
3. pH = 11.7 5. 6.07 × 105
4. 2.43 × 10–3gm 7. 4.19 × 10–3 mole L–1
5. KC= 3.08 × 10–2, 0.123 M 9. 5 × 10–13, % of Ag+= 99.83%
6. 1.7 × 10–3M 11. 7.68 × 10–4 mole L–1
7. 0.0135 mole L–1 13. pH = 4
8. [S2–] = 6.8 × 10–2 M 15. pH = 2.392

 

13. Answers to Objective Assignments 

 

LEVEL – I

 

1. A 2. A
2. A 4. B
3. A 6. D
4. C 8. B
5. B 10. D
6. A 12. B
7. D 14. A
8. A 

LEVEL – II

1. A 2. B
2. D 4. B
3. B 6. D
4. A 8. A
5. B 10. A
6. B 12. A
7. A 14. B
8. B

6