Question

30-32) What is the pH of the resulting solution when 5.0 mL of a 0.0525 M HCI solution is mixed with 50.0 mL of a 0.0125 M HC2H302 solution? Ka for HC2H302 is 1.8 x 105

Answer 1

Dissociation equillibrium of HC2H3O2 is

HC2H3O2 <------> C2H3O2^- + H⁺

Ka = [C2H3O2^-][H⁺] /[HC2H3O2]

Number of moles of HCl = (0.0525mol/1000ml)× 5ml = 0.0002625

Total volume after mixing = 55ml

[HCl] = (0.0002625mol/55ml)×1000ml = 0.004773M

Number of moles of HC2H3O2 = (0.0125mol/1000ml)×50.0ml = 0.000625mol

[HC2H3O2] = (0.000625mol/55ml)×1000ml = 0.01136M

Initial concentration

[HC2H3O2] = 0.01136M

[C2H3O2^-] = 0

[H⁺] = 0.004773M

change in concentration

[HC2H3O2] = -x

[C2H3O2^-] = +x

[H⁺] = +x

Equillibrium concentration

[HC2H3O2] = 0.01136 -x

[C2H3O2^-] = x

[H⁺] = 0.004773 + x

x(0.004773 +x) /(0.01136 -x) = 1.8×10^-5

solving for x

x = 0.00004231

[H⁺ ] = 0.004773 + 0.00004231 = 0.004815

pH = -log[H⁺]

pH = - log(0.004815)

pH = 2.32