Question

Calculate the ionization constant of a weak acid.

The values given are 0.5 M of hydrogen cyanide and 0.0068 M for hydronium and cyanide ions.

$HCN\left(aq\right)+H_{2}O\to H_3{O}^{+}+C{N}^{-}$

• A. $7.9\times {10}^{-5}$
• B. $9.2\times {10}^{-5}$
• C. $1.3\times {10}^{-5}$
• D. $2\times {10}^{-5}$

Answer 1

Answer: (B)

$9.2\times {10}^{-5}$

Given,

$\left[HCN\right]$ is $0.5M$
$\left[H_3{O}^{+}\right]$ is $0.0068M$
$\left[C{N}^{-}\right]$ is $0.0068M$
We can assume here that, $\left[H3O^{+}\right]=\left[C{N}^{-}\right]$
The equilibrium constant would be as follows:
$K_a=\frac{\left[H_3{O}^{+}\right]\left[C{N}^{-}\right]}{\left[HCN\right]}=\frac{{0.0068}^2}{0.5}$ $=9.2\times {10}^{-5}$