how to do this
What is the pH of a 1.0 mol L-1 NaCN aqueous solution? (Ka of HCN 6.2 x 10-10) A24. A) 2.4
B) 3.4 C) 9.2 D) 8.8 E) 11.6
Solution
NaCN(aq) ----------------> Na^+ (aq) + CN^- (aq)
1M 1M
CN^- (aq) + H2O ---------------> HCN(aq) + OH^- (aq)
I 1 0 0
C -x +x +x
E 1-x +x +x
Kb = Kw/Ka
= 1*10^-14/6.2*10^-10 = 1.6*10^-5
Kb = [HCN][OH^-]/[CN^-]
1.6*10^-5 = x*x/1-x
1.6*10^-5 *(1-x) = x^2
x = 0.004
[OH^-] = x = 0.004M
POH = -log[OH^-]
= -log0.004

= 2.3979
PH = 14-POH
= 14-2.3979 = 11.6
E. 11.6