Question

Consider the titration of 100 ml of 1M NaF with 1M HCl. Calculate the initial pH.

Answer 1

F^- partly hydrolysed by water

F^-(aq) + H2O(l) < ------- > HF(aq) + OH^-(aq)

K_b = [HF][OH^-] / [F^-]

K_b = K_w/K_a

where,

K_w = ionic product of water , 1.00 × 10^-14

K_a = K_a of HF , 6.92 × 10^-4

K_b = 1.00 ×10^-14/ 6.92 ×10^-4

K_b = 1.45 × 10^-11

Initial concentration

[F^-] = 1

[HF] = 0

[OH^-] = 0

change in concentration

[F^-] = -x

[HF] = +x

[OH^-] = + x

Equilibrium concentration

[F^-] = 1- x

[HF] = x

[OH^-] = x

so,

x²/(1 - x) = 6.92 ×10^-4

solving for x

x = 0.02596

[OH^-] = 0.02596M

pOH = -log[OH^-]

pOH = -log(0.02596)

pOH = 1.59

pH = 14 - pOH

pH = 14 - 1.59

pH = 12.41

Therfore

Initial pH = 12.41