what is the pH of a 0.35M KCN solution? K a for HCN=6.20x10^-10.?

Answer 1

The reaction we follow is:
CN- +H2O -> HCN + OH-
The Ka reaction is
[CN-][H+]/[HCN]= Ka, which is not quite what we have. We can "flip" this to get
[HCN]/[CN-][H+]=1/ Ka, and realizing that we can multiply both sides by Kw=[H+][OH-] to get
[HCN][OH-]/[CN-]= Kw/Ka.
I show this because this is the general means of attack if you start with a salt of a weak acid or a weak base. The initial solution is 0.35M in CN-. A small amount, x, reacts to form HCN, and in doing so forms x moles of [OH-]. Probably, x<< 0.35M. The math is
x^2 / 0.35 = 10x10-15/6.2x10-10
x^2 / 0.35 = 1.65 x 10-5 appx
x^2 = 6x10-6 appx
x= 2.4 x 10-3
Note, the assumption of x vs. 0.35 looks justified.
We have [OH-], and as a "short cut", we can find
pOH= 3-log 2.4. This is a pOH of 2.6. From the definition of Kw, pOH + pH =14. Then pH= 11.4.