The PH of an aqueous solution of NaCN is 10.38. What is [CN-] in this solution?
Sorry for double posting but I got a solution of [CN-] = 0.0036 M. Can someone verify if that is right.
Using 6.2 x 10^-10 for Ka for HCN, I obtained 0.00357 M for (CN^-) which rounds to 0.0036. I think you probably worked it correctly.