Question

A sample of 0.105 M CH3CO2H is titrated with 0.125 M KOH. What is the appropriate pH of the resulting solution when the equivalence point is reached? Ka= 1.8 x 10^-5

Answer 1

in equivalence

CH3CO2H + KOH = CH3CO2K + H2O

CH3CO2K -- > CH3CO2-

CH3CO2- + H2O <--< CH3CO2H

expect hydrolysis

CH2O-(aq) + H2O(l) <->CH2OH + OH-(aq)

Let HA --> CH2OH and A- = CH2O- for simplicity

since A- is formed

the next equilibrium is formed, the conjugate acid and water

A- + H2O <-> HA + OH-

The equilibrium s best described by Kb, the base constant

Kb by definition since it is an base:

Kb = [HA ][OH-]/[A-]

Ka can be calculated as follows:

Kb = Kw/Ka = (10^-14)/(1.8*10^-4) = 5.55*10^-11

get ICE table:

Initially

[OH-] = 0

[HA] = 0

[A-] = M

the Change

[OH-] = + x

[HA] = + x

[A-] = - x

in Equilibrium

[OH-] = 0 + x

[HA] = 0 + x

[A-] = M - x

substitute in Kb expression

Kb = [HA ][OH-]/[A-]

M = 0.105/2 = 0.05

5.55*10^-10 = x*x/(0.05-x)

solve for x

x^2 + Kb*x - M*Kb = 0

solve for x with quadratic equation

x = OH- =5.27*10^-6

[OH-]  =5.27*10^-6

pOH = -log(OH-) = -log(5.27*10^-6= 5.28

pH = 14-5.28= 8.72

pH = 8.72