Question

A monoprotic acid, HA, is dissolved in water: The equilibrium concentrations of the reactants and products are [HA] = 0.160 M [H ] = 4.00

Answer 1

A monoprotic acid dissociates according to the reaction equation:
HA(aq) ------> H+(aq) + A-(aq)

The equilibrium concentrations in M satisfy the relation:
Ka = [H+]·[A-] / [HA]

For this acid
Ka = (4.00×10^-4 * 4.00×10^-4 )/ 0.16 M
Ka= 10^-6

Hence,
pKa = - log10(Ka) = - log10(10^-6) = 6

Answer 2

A monoprotic acid dissociates according to the reaction equation:
HA(aq) ? H?(aq) + A?(aq)

The equilibrium concentrations in M satisfy the relation:
Ka = [H?]·[A?] / [HA]

For this acid
Ka = 2.00×10?4 · 2.00×10?4 / 0.26 M
= 1.54×10?7

Hence,
pKa = - log10(Ka) = - log10(1.54×10?7) = 6.81