# Answer on Inorganic Chemistry Question for ston

Question #25158

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?

Expert's answer

The chemical equation for the dissociation reaction

HA <=> H

[HA] [H

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

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

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%

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 a^{2}/ (1-a)a << 1, that,s why we can write K(dis) = C x a

^{2}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%

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