Question

A weak acid HA is 1.0% dissociated in a 1.0 M solution. What would its percent dissociation be in a 5.0 M solution?

Answer 1

The chemical equation for the dissociation reaction

HA <=> H⁺ + A^-

[HA] [H⁺] [A^-]
Concentration
before dissociation C 0 0


Concentration after C x (1-a) C x a C x a
dissociation


C - concentration (M)
a - dissociation percent

The equation for dissociation constant:


K(dis) = [H⁺] x [A^-]/[HA] = (C x a) x (C x a) / (C x (1-a)) = C x a² / (1-a)

a << 1, that,s why we can write K(dis) = C x a²

K(dis) = 1.0 x 0.01 x 0.01 = 1.0 x 10^(-4)

a = (K/c)^0.5 x 100%

In a 5.0 M solution the dissociation percent is

a = (1.0 x 10^(-4) / 5.0)^0.5 x 100% = 0.45 %

Answer : a = 0.45%