Chemistry Common Ion Effect, Hydrolysis of Salts and the pH of their Solutions and Buffer solutions
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Class-11 chapter-7 EQUILIBRIUM

Common Ion Effect, Hydrolysis of Salts and the pH of their Solutions and Buffer solutions

Topic to be covered

`star` Common Ion Effect in the Ionization of Acids and Bases.
`star` Hydrolysis of Salts and the pH of their Solutions.
`star` Buffer solutions.

Common Ion Effect in the Ionization of Acids and Bases

`=>` Consider an example of acetic acid dissociation equilibrium represented as:

`color{red}(CH_3COOH(aq) ⇌ H+(aq) + CH_3COO^– (aq))`

or `color{red}(HAc(aq) ⇌ H^+ (aq) + Ac^– (aq))`

`color{red}(K_a = [H^+][Ac^– ] // [HAc])`

`=>` Addition of acetate ions to an acetic acid solution results in decreasing the concentration of hydrogen ions, `color{red}([H^+])`. Also, if `color{red}(H^+)` ions are added from an external source then the equilibrium moves in the direction of undissociated acetic acid i.e., in a direction of reducing the concentration of hydrogen ions, `color{red}([H^+])`. This phenomenon is an example of common ion effect.

`color{purple}(✓✓)color{purple} " DEFINITION ALERT"`

`color{green}("COMMON ION EFFECT")` : It can be defined as a shift in equilibrium on adding a substance that provides more of an ionic species already present in the dissociation equilibrium. Thus, we can say that common ion effect is a phenomenon based on the Le Chatelier’s principle.

`=>` In order to evaluate the `color{red}(pH)` of the solution resulting on addition of `0.05M` acetate ion to `0.05M` acetic acid solution, we shall consider the acetic acid dissociation equilibrium once again,

`color{red}(HAc(aq) ⇌ H^+(aq) + Ac^–(aq))`

`color{green}("Initial concentration (M)")`

`color{red}(0.05 \ \ \ \ \ \ \ \ 0 \ \ \ \ \ \ \ 0.05)`

Let `color{red}(x )`be the extent of ionization of acetic acid.

`color{green}("Change in concentration (M)")`

`color{red}(-x \ \ \ \ \ \ \ \ + x \ \ \ \ \ \ \ \ +x)`

`color{green}("Equilibrium concentration (M)")`

`color{red}(0.05 -x \ \ \ \ \ \ \ \ x \ \ \ \ \ \ \ 0.05+x)`

Therefore,
`color{red}(K_a= [H^+][Ac^– ]//[H Ac] = {(0.05+x)(x)}/(0.05-x))`

As `color{red}(K_a)` is small for a very weak acid, `color{red}(x < < 0.05).`

Hence, `color{red}((0.05 + x) ≈ (0.05 – x) ≈ 0.05)`

Thus,
`color{red}(1.8 × 10^(–5) = (x) (0.05 + x) // (0.05 – x))`

`color{red}(= x(0.05) // (0.05) = x = [H^+] = 1.8 × 10^(–5)M)`

`color{red}(pH = – log(1.8 × 10^(–5)) = 4.74)`



Q 3150112014

Calculate the `pH` of a `0.10M` ammonia solution. Calculate the `pH` after `50.0 mL` of this solution is treated with `25.0 mL` of `0.10M HCl.` The dissociation constant of ammonia, `K_b = 1.77 × 10^(–5)`

Solution:

`NH_3 + H_2O → NH_4^+ + OH^–`


`K_b = [NH_4^+][OH^–] // [NH_3] = 1.77 × 10^(–5)`

Before neutralization, `[NH_4^+] = [OH^–] = x [NH_3] = 0.10 – x approx 0.10`

`x^2 // 0.10 = 1.77 × 10^(–5)`

Thus, `x = 1.33 × 10^(–3) = [OH^–]`
Therefore,`[H^+] = K_w // [OH^–] = 10^(–14) // (1.33 × 10^(–3)) = 7.51 × 10^(–12)`
`pH = –log(7.5 × 10^(–12)) = 11.12`
On addition of `25 mL` of `0.1M HCl`
solution (i.e., 2.5 mmol of `HCl`) to 50 mL of 0.1M ammonia solution (i.e., 5 mmol of `NH_3`), 2.5 mmol of ammonia molecules
are neutralized. The resulting 75 mL solution contains the remaining unneutralized 2.5 mmol of `NH_3` molecules and 2.5 mmol of `NH_4^+`.

`tt (( NH_3 , + , HCl , → , NH_4^+ , + , Cl^(-) ) , ( 2.5 , , 2.5 , , 0 , , 0 ),("At equilibrium " , , , , , , ), (0 , , 0 , , 2.5 , , 2.5))`

The resulting `75 mL` of solution contains `2.5 mmol` of `NH_4^+` ions (i.e., 0.033 M) and 2.5 mmol (i.e., 0.033 M ) of uneutralised NH3 molecules. This `NH_3` exists in the following equilibrium:

`tt (( NH_4OH , ⇌ , NH_4^+ , + , OH^- ) , (0.033M-y , , y , , y))`

where, `y = [OH^–] = [NH_4^+]`

The final 75 mL solution after neutralisation already contains 2.5 m mol `NH_4^+` ions (i.e. 0.033M), thus total concentration of `NH_4^+` ions is given as:

`[NH_4^+ ] = 0.033 + y`

As `y` is small,` [NH_4OH] approx 0.033 M` and `[NH_4^+] approx 0.033M.`

We know,

`K_b = [ NH_4^+] [ OH^(-) ] // [NH_4OH]`

`y(0.033)//(0.033) = 1.77 xx 10^(-5) M`

Thus, `y = 1.77 × 10^(–5) = [OH^–]`

`[H^+] = 10^(–14) // 1.77 × 10^(–5) = 0.56 × 10^(–9)`
Hence, `pH = 9.24`

Hydrolysis of Salts and the pH of their Solutions

`=>` Salts formed by the reactions between acids and bases in definite proportions, undergo ionization in water.

`=>` The cations/anions formed on ionization of salts either exist as hydrated ions in aqueous solutions or interact with water to reform corresponding acids/bases depending upon the nature of salts. The later process of interaction between water and cations/anions or both of salts is called hydrolysis. The pH of the solution gets affected by this interaction. The cations (e.g., `color{red}(Na^+, K^+, Ca^(2+), Ba^(2+))`, etc.) of strong bases and anions (e.g., `color{red}(Cl^–, Br^–, NO_3^–, ClO_4^–)` etc.) of strong acids simply get hydrated but do not hydrolyse, and therefore the solutions of salts formed from strong acids and bases are neutral i.e., their `color{red}(pH)` is `7`. However, the other category of salts do undergo hydrolysis.

`color{green}("We now consider the hydrolysis of the salts of the following types :")`

(i) salts of weak acid and strong base e.g., `color{red}(CH_3COONa).`
(ii) salts of strong acid and weak base e.g., `color{red}(NH_4Cl)`, and
(iii) salts of weak acid and weak base, e.g., `color{red}(CH_3COONH_4)`.

`=>` In the first case, `color{red}(CH_3COONa)` being a salt of weak acid, `color{red}(CH_3COOH)` and strong base, `color{red}(NaOH)` gets completely ionised in aqueous solution. `color{red}(CH_3COONa(aq) → CH_3COO^– (aq)+ Na^+(aq))`

Acetate ion thus formed undergoes hydrolysis in water to give acetic acid and `color{red}(OH^–)` ions

`color{red}(CH_3COO^–(aq)+H_2O(l) ⇌ CH_3COOH(aq)+OH^–(aq))`

Acetic acid being a weak acid `color{red}((K_a = 1.8 × 10^(–5)))` remains mainly unionised in solution. This results in increase of `color{red}(OH^(–))` ion concentration in solution making it alkaline. The `color{red}(pH)` of such a solution is more than 7.

`=>` Similarly, `color{red}(NH_4Cl)` formed from weak base, `color{red}(NH_4OH)` and strong acid, `color{red}(HCl)`, in water dissociates completely.

`color{red}(NH_4Cl(aq) → NH_4^+ (aq) +Cl^– (aq))`

Ammonium ions undergo hydrolysis with water to form `color{red}(NH_4OH)` and `color{red}(H^+)` ions

`color{red}(NH_4^+ (aq) + H_2O (l) ⇌ NH_4 OH (aq) +H^+ (aq))`

Ammonium hydroxide is a weak base `(color{red}(K_b = 1.77 × 10^(–5)))` and therefore remains almost unionised in solution. This results in increased of `color{red}(H^+)` ion concentration in solution making the solution acidic. Thus, the `color{red}(pH)` of `color{red}(NH_4Cl)` solution in water is less than 7.

`=>` Consider the hydrolysis of `color{red}(CH_3COONH_4)` salt formed from weak acid and weak base. The ions formed undergo hydrolysis as follow:

`color{red}(CH_3COO^– + NH_4^+ + H_2O ⇌ CH_3COOH + NH_4OH)`


`color{red}(CH_3COOH)` and `color{red}(NH_4OH)`, also remain into partially dissociated form :


`color{red}(CH_3COOH ⇌ CH_3COO^(–) + H^+)`


`color{red}(NH_4OH ⇌ NH_4^+ + OH^–)`


`color{red}(H_2O ⇌ H^(+) + OH^(–))`

Without going into detailed calculation, it can be said that degree of hydrolysis is independent of concentration of solution, and
`color{red}(pH)` of such solutions is determined by their `color{red}(pK)` values:

`color{red}(pH = 7 + ½ (pK_a – pK_b))` ............. (7.38)


The `color{red}(pH)` of solution can be greater than 7, if the difference is positive and it will be less than 7, if the difference is negative.

Q 3110212110

The `pK_a` of acetic acid and `pK_b` of ammonium hydroxide are 4.76 and 4.75 respectively. Calculate the `pH` of ammonium acetate solution.

Solution:

`pH = 7 + ½ [pK_a – pK_b]`

`= 7 + ½ [4.76 – 4.75]`

`= 7 + ½ [0.01] = 7 + 0.005 = 7.005`

BUFFER SOLUTIONS

`=>` Many body fluids e.g., blood or urine have definite `color{red}(pH)` and any deviation in their `color{red}(pH)` indicates malfunctioning of the body.

`=>` The control of `color{red}(pH)` is also very important in many chemical and biochemical processes.

`=>` Many medical and cosmetic formulations require that these be kept and administered at a particular `color{red}(pH)`.

`color{purple}(✓✓)color{purple} " DEFINITION ALERT"`
The solutions which resist change in `color{red}(pH)` on dilution or with the addition of small amounts of acid or alkali are called Buffer Solutions.

Buffer solutions of known `color{red}(pH)` can be prepared from the knowledge of `color{red}(pK_a)` of the acid or `color{red}(pK_b)` of base and by controlling the ratio of the salt and acid or salt and base. A mixture of acetic acid and sodium acetate acts as buffer solution around `color{red}(pH 4.75)` and a mixture of ammonium chloride and ammonium hydroxide acts as a buffer around `color{red}(pH 9.25).`

 
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