Question

Use the followign information to calculate the pH of solution A:

(a) Solution A is a 0.100< solution of the weak monoprotic acid HX at 25 degrees C.

(b) soluton B is a 0.500 M solution of the salt NaX. It has a pH of 10.02 at 25 degrees C.

The answer: pH of A = 4.17

Answer 1

To find pH of solution A we nned to know Ka of weak monoprotic acid Ka which we can find out from it's salt solution NaX i.e. solution B.

a) For Solution B:

NaX is a strong electrolyte and ionizes completely as,

NaX ------> Na⁺ (aq) + X^- (aq).

So, [X^-] = [NaX] = 0.5 M (given)

X^- undergo hydrolysis and gives HX and HO^- ions as per Kb of A^- conjuagate base.

X^- (aq) + H2O <-------> HX + HO^- (aq)

Kb = [HX][HO^-] / [X^-]. ------------- (1)

We have, pH = 10.02 (for salt solution given)

pOH + pH = 14

pOH = 14 - pH = 14 - 10.02 = 3.98

Then by definition of pOH,

pOH = -log[HO^-]

[HO^-] = 10^-pOH. = 10^-3.98 = 1.05 x 10^-4 M

Stoichiometry of hydrolysis says,

[HX] = [HO^-] =1.05 x 10^-4 M.

And [X^-] = 0.5 M using these values in eq.(1)

Kb = (1.05x10^-4)(1.05x10^-4) / 0.5

Kb = 2.2 x 10^-8.

We know that for aqueous solutions,

Ka x Kb = Kw

Ka x 2.2 x 10^-8 = 1.00 x 10^-14.

Ka = 1.00 x 10^-14 / 2.2 x 10^-8.

Ka = 4.55 x 10^-7.

This is the dissociation constant for conjugated acid HX (HX is conjugated acid for X^- base).

This Ka value we will use in pH determination of solution A.

b) For solution A,

HX a weak acid ionizes as,

HX <------------> H⁺ (aq) + X^- (aq).

Ka = [H⁺][X^-] / [HX] = 4.55 x 10^-7. --------- (2)

Initially, [HX] = 0.1 M, let at equilibrium 'C' M HX ionizes and hence ICE table is given as,

HX <------------> H⁺ (aq) + X^- (aq).

Initially 0.1 0 0

Change -C +C +C

At eq^m (0.1-C) C C

Using equilibrium concentration in eq.(2)) we get,

(C) (C) / (0.1-C) = 4.55 x 10^-7.

HX being weak acid Small concentration assumption holds so 0.1 - C = 0.1 (apprx.)

Hence we write,

C² = 0.1 x 4.55 x 10^-7 .

C² = 4.55 x 10^-8.

C = 2.13 x 10^-4.

By ICE table,

[H⁺] = C = 2.13 x 10^-4 M

By definition of pH,

pH = -log[H⁺]

pH = -log(2.13 x 10^-4)

pH = 3.67

pH of solution A is 3.67

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