Question

Chloracetic acid is a stronger monoprotic acid than acetic acid. In a 0.10M solution of this acid, the pH was 1.96. Find the Ka. (Hint: you can use the general formula HA to represent chloroacetic acid)

Answer 1

Equilibrium reaction

$HC_2H_2O_2Cl+H_2O->C_2H_2O_2Cl^-H_3O^+$

$Ka=\frac{products}{reactants}$

$Ka=\frac{[C_2H_2O_2CL^-][H_3O^+]}{[HC_2H_2O_2CL]}$

$Ka=\frac{[X][X]}{0.10-X}$

$Ka=\frac{X^2}{0.10-X}$

we can calculate the value of X using [H₃O⁺] and the given pH of the solution

$[H^+]=[H_3O^+]$

$H_3O^+=10^{-pH}$

$H_3O^+=10^{-1.96}$

$H_3O^+=X=0.010965$

Hence

$Ka=\frac{X^2}{0.10-X}$

$Ka=\frac{[0.010965]^2}{0.10-[0.010956]}=1.35\times{10^{-3}}$