Question

1. Write the acid dissociation reaction for following solutions, and calculate the pH for each of them: a) 0.001 M HCl b) 0.005 M H2SO4 c) 0.005 M KOH

Answer 1

(a)

HCl(aq) ------> H+ (aq) + Cl- (aq)

[H+] = [HCl] = 0.001 M

pH = - log[H+] = - log(0.001) = 3

(b)

H2SO4(aq) ------> H+(aq) + HSO4- (aq)

HSO4- (aq) --------> H+ (aq) + SO4^2- (aq)

For 1st reaction

[H2SO4] = [H+] = 0.005 M

But we have to consider the dissociation of HSO4- as incomplete because there is no mention of complete dissociation.

HSO4- --------> H+ + SO4^2-

0.005                 0             0      (Initial)

0.005 - x             x             x      (Final)

Ka of HSO4- = [H+] [SO4^2-] / [HSO4-]

0.012 = x^2 / (0.005 - x)

x^2 + 0.012 x - 6 * 10^-5 = 0

x = 0.0038 M = [H+]

SO,

Total [H+] = 0.005 + 0.0038 = 0.0088 M

pH = -log[H+] = -log(0.0088) = 2.056

(c)

KOH (aq) -------> K+ (aq) + OH- (aq)

[OH-] = 0.005 M

pOH = -log[OH-] = -log(0.005)

pOH = 2.301

pH + pOH = 14

pH = 14 - pOH = 14 - 2.301

pH = 11.699