Question

Determine the pH of each of the following two-component solutions.

1) 6.0×10?2 M KOH and 2.0×10?2 M Ba(OH)2

2) 0.270 M NH4NO3 and 0.103 M HCN

3) 7.5×10^?2 M RbOH and 0.130 M NaHCO3

4) 8.8×10^?2 M HClO4 and 2.0×10^?2 M KOH

5) 0.115 M NaClO and 5.50×10^?2 M KI

Answer 1

pH of solution

1) Total [OH-] = 6 x 10^-2 + 2 x 2 x 10^-2 = 0.1 M

pOH = -log[OH-] = -log(0.1) = 1

pH = 14 - pOH = 14 - 1 = 13

2) [NH4+] = 0.270 M

NH4+ <==> NH3 + H+

Ka = 5.55 x 10^-10 = x^2/0.270

[H+] = 1.22 x 10^-5 M

[HCN] = 0.103 M

HCN <==> H+ + CN-

Ka = 6.2 x 10^-10 = x^2/0.103

[H+] = x = 8 x 10^-6 M

Total [H+] = 2.02 x 10^-5 M

pH = -log(2.02 x 10^-5) = 4.70

3) RbOH strong base

[OH-] = 7.5 x 10^-2 M

pOH = -log(7.5 x 10^-2) = 1.12

pH = 14 - 1.12 = 12.88

4) excess [HClO4] = [H+] = 8.8 x 10^-2 - 2 x 10^-2 = 6.8 x 10^-2 M

pH = -log(6.8 x 10^-2) = 1.17

5) [ClO-] = 0.115 M

ClO- + H2O <==> HClO + OH-

Kb = 3.45 x 10^-7 = x^2/0.115

x = [OH-] = 2 x 10^-4 M

pOH = -log[OH-] = 3.70

pH = 14 - 3.70 = 10.3