The pH is determined by the hydrolysis of the CN^- as follows:
.....................CN^- + HOH ==>HCN + OH^-
I..................0.747.......................0..........0
C...................-x...........................x...........x
E.................0.747-x.....................x...........x
Kb (for CN^-) = Kw/Ka(for HCN) = (HCN)(OH^-)/(CN^-)
Plug the E line into the Kb expression and solve for x = (OH^-). Convert to (H^+) and pH. Post your work if you get stuck.
What is the PH of a 0.747M sodium cyanide solution (NaCN) knowing that the Ka of Hydrocyanic acid is 4.9x10-10?
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