Answer to Question #25158 in Inorganic Chemistry for ston
2013-02-25T08:45:58-05:00
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?
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2013-02-27T06:03:19-0500
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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