Question

a. Calculate the pH of a 0.538 M aqueous solution of hydrofluoric acid (HF, K_a = 7.2×10^-4).

b. Calculate the pH of a 0.0242 M aqueous solution of nitrous acid (HNO₂, K_a = 4.5×10^-4).

Answer 1

a) acid dissociation eq is given by

HF (aq) <---> H+ (aq) + F- (aq)

at equilibirum [HF] = 0.538-X , [H+] = [F-] = X

Ka = [H+] [F-] / [HF]

7.2 x 10^ -4 = ( X^2) / ( 0.538-X)

( here we do approximation 0.538-X as 0.538 , since [HF] is not very low but we get low value of X as Ka is small)

now 7.2 x 10^ -4 = X^2 / ( 0.538)

X = [H+] = 0.0197

pH = -log ( 0.0197) = 1.7

b) HNO2 (aq) <---> H+ (aq) + NO2-(aq)

the Ka expression is same , where at equlibrium

[HNO2] = 0.0242-X , [H+] =[NO2-] = X

4.5 x 10^ -4 = X^2 / ( 0.0242-X)

X^2 + ( 4.5 x 10^ -4) X - 0.000011 = 0

X = 0.00308 = [H+]

pH = -log ( 0.00308)

= 2.5