Question

The pH of the $0.10M$ hydrocyanic acid solutions is 5.2. What will be the value of $K_a$ for hydrocyanic acid ?

• A. $2.0\times {10}^{-5}$
• B. $4.0\times {10}^{-5}$
• C. $2.0\times {10}^{-10}$
• D. $4.0\times {10}^{-10}$

Answer 1

The correct option is D $4.0\times {10}^{-10}$

Given,
pH of hydrocyanic acid is 5.2
$pH=-\text{log}{H}^{+}=5.2$
$\text{log}\left[H^{+}\right]=-5.2$
$\left[H^{+}\right]=\text{Antilog}(-5.2)=6.31\times {10}^{-6}M$
$HCN\rightleftharpoons H^{+}+C{N}^{-}$
$0.1-6.31\times {10}^{-6}6.31\times {10}^{-6}M6.31\times {10}^{-6}M$
$=0.1M$ ($6.31\times {10}^{-6}$ is too smaller than 0.1, so we can neglect it)
$K_a=\frac{\left[H^{+}\right]\left[C{N}^{-}\right]}{\left[HCN\right]}$
$=\frac{(6.31\times {10}^{-6}M)\times (6.31\times {10}^{-6})M}{0.1M}$
$=3.9816\times {10}^{-10}=4.0\times {10}^{-10}$