Ionic EquilibriumMedium
Question
What is the pH of a 0.50 M aqueous NaCN solution? The value of pKb of CN– is 4.70 (log 2 = 0.3).
Options
A.3.5
B.11.5
C.4.7
D.9.3
Solution
$P^{K_{b}}\text{ of C}\text{N}^{-} = 4.70 \Rightarrow P^{K_{a}}\text{ of HCN = 9.30}$
Now, $P^{H} = 7 + \frac{1}{2}\left( P^{K_{a}} + \log C \right) = 7 + \frac{1}{2}\left( 9.30 + \log 0.5 \right) = 11.5$
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