Question

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

0.260 M NH4NO3 and 0.102 M HCN

7.5×10?2 M RbOH and 0.100 M NaHCO3

9.2×10?2 M HClO4 and 2.0×10?2 M KOH

0.120 M NaClO and 5.00×10?2 M KI

Answer 1

pH of mixture

0.260 M NH4NO3 and 0.102 M NaCN

NH4+ + H2O <==> NH3 + H3O+

Ka = 5.55 x 10^-10 = x^2/0.260

x = [H3O+ = 1.20 x 10^-5 M

CN- + H2O <==> HCN + OH-

Kb = 1.61 x 10^-5 = x^2/0.102

x = [OH-] = 1.30 x 10^-3

[H3O+] = 1 x 10^-14/1.30 x 10^-3 = 7.80 x 10^-12 M

Total [H3O+] = 1.20 x 10^-5 + 7.80 x 10^-12 = 1.20 x 10^-5 M

pH = -log[H3O+] = 4.92

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RbOH is a strong base

pH dependent only on RbOH

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

[H3O+] = 1 x 10^-14/7.5 x 10^-2 = 1.33 x 10^-13 M

pH = -log[H3O+] = 12.90

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HClO4 + KOH --> KClO4 + H2O

2 x 10^-2 M KOH will neutralize 2 x 10^-2 M HClO4

remaining [HClO4] = 9.2 x 10^-2 - 2 x 10^-2 = 0.072 M

pH = -log(0.072) = 1.14

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pH dependent only on ClO- concentration

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

Kb = 3.45 x 10^-7 = x^2/0.120

x = [OH-] = 2.03 x 10^-4 M

[H3O+] = 1 x 10^-14/2.03 x 10^-4 = 4.91 x 10^-11 M

pH = -log(4.91 x 10^-11) = 10.31