Question

Question: The solubility of a salt in water increases if either ion of the salt undergoes hydrolysis. Comput the molar solubility of AgCN in water assuming no hydrolysis. B) compute the molar solubility of AnCN in water assuming hydrolysis occurs.

The equation would be AgCN goes to Ag + CN. You would get S^2= Ksp value. Is this right? I'm not sure i'm setting this up right. How do you solve when hydrolysis occurs in the system?

Answer 1

Right, S = sqrt Ksp for no hydrolysis.
With hydrolysis, here is what you have.
AgCN ==> Ag+ + CN^-
and CN^- + HOH ==> HCN + OH^-
Let S = solubility AgCN, then (Ag^+) = S and total (Ag^+) = (CN^-) + (HCN)
so S = (CN^-) + (HCN).
For HCN we can write a Ka,
Ka = (H^+)(CN^-)/(HCN)
Since the problem doesn't list a H^+ (which would make it even more soluble), I would use 1 x 10^-7 for (H^+) in the Ka expression and solve for
(HCN). Check my work but I believe
(HCN) = 1 x 10^-7*(CN^-)/Ka.
Look up Ka and (HCN) = some number*(CN^-). Then
S = (CN^-) + (HCN from above) and solve for (CN^-) in terms of S.
Then use Ksp expression to solve for S.
Using 7.2 x 10^-11 for Ka for HCN, I obtain 8.5 x 10^-6 for solubility with no hydrolysis and 5.9 x 10^-5 for solubility with hydrolysis (or its about 10x more soluble with hydrolysis).

Answer 2

I couldn't find a ksp value for the AgCn. How do you know to use a Ka value

Answer 3

I made a typo.
Ka for HCN = 2.1 x 10^-9
and Ksp for AgCN = 7.2 x 10^-11 according to my quant book (about 15 years old). However, the final values i quoted still are ok; i.e., I used the correct values in my calculations but I didn't copy them correctly when I posted my response. Are you asking how I know to use a Ka for HCN. Because HCN is a weak acid. That's why CN^- hydrolyzes.