Question

Hi

"Ka for HCN is 4.9 x 10^-10. What is the pH of a 0.068 M aqueous solution of sodium cyanide?"

How do I do this question?

Thank you very much

Answer 1

This is a hydrolysis question. The CN^- ion is base and it reacts with water as follows:
CN^- + HOH ==> HCN + OH^-

Kb = (Kw/Ka) = (HCN)(OH^-)/(CN^-)

X = HCN
X = OH
CN is given in the problem. Kw you know, Ka is given in the problem. Solve for X, convert OH to pOH, then to pH.