Question

0.29mole of an unknown monoprotic acid is dissolved in enough water to make 1.55L of a solution with a pH 3.82. What is the dissociation constant of the acid?

Answer 1

Any general acid dissociates like so: \(HA→H^++A^−\); with \(K=\dfrac{[H^+][A^−]}{[HA]}\)

we are given: \([H^+]=10^{−pH} mol/L=[A^−]\)

also, \([HA]=\dfrac{0.29 ~mol}{1.55~ L}=0.187~mol/L\)

\(K=\dfrac{(10^{−pH})^2}{(0.187 mol/L−10^{−pH})}\)