Ionic EquilibriumMedium
Question
A volume of 20 ml of 0.8 M-HCN solution is mixed with 80 ml of 0.4 M-NaCN solution. Calculate the pH of the resulting solution. The value of Ka of HCN = 2.5 × 10−10 (log 2 = 0.3).
Options
A.9.9
B.9.3
C.4.1
D.4.7
Solution
$P^{H} = P^{K_{a}} + \log\frac{\left\lbrack CN^{-} \right\rbrack_{0}}{\lbrack HCN\rbrack_{0}}$
$= - \log\left( 2.5 \times 10^{- 10} \right) + \log\frac{80 \times 0.4/100}{20 \times 0.8/100} = 9.9$
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