Question

Calculate : $K_a$ for $HCN$ given $K_b\left({CN}^{-}\right)=2.5\times {10}^{-5}$

• A. ${10}^{-5}$
• B. $2\times {10}^{-7}$
• C. $4\times {10}^{-10}$
• D. None of these

Answer 1

The correct option is C $4\times {10}^{-10}$

$HCN\rightleftharpoons H^{+}+C{N}^{-}$......(1)
$C{N}^{-}+H_{2}O\rightleftharpoons HCN+O{H}^{-}{K}_b=2.5\times {10}^{-5}$ ......(2)
$H_{2}O\rightleftharpoons H^{+}+O{H}^{-}{K}_w=1\times {10}^{-14}$......(3)
Substract equations (3) from (2), we get
$C{N}^{-}+O{H}^{-}\rightleftharpoons HCN\frac{K_b}{K_w}=\frac{1}{K_a}$
$\frac{2.5\times {10}^{-5}}{1\times {10}^{-14}}=\frac{1}{K_a}$
$K_a=4\times {10}^{-10}$