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Answer to Question #25158 in Inorganic Chemistry 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+ + 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%

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