Question

What is the hydrolysis constant of the $OC{l}^{-}$ ion?

Given the ionization constant of $HOCl$ is $3.0\times {10}^{-8}$

• A. $3.33\times {10}^{-8}$
• B. $3.33\times {10}^{-7}$
• C. $3.0\times {10}^{-7}$
• D. $3.33\times {10}^{-6}$

Answer 1

The correct option is B $3.33\times {10}^{-7}$

The ionization constant of $HOCl$ is $3.0\times {10}^{-8}$

$HOCl+H_{2}O\to H_3{O}^{+}+OC{l}^{-}$

$K_a=\frac{\left[H_3{O}^{+}\right]\left[OC{l}^{-}\right]}{\left[HOCl\right]}=3.0\times {10}^{-8}$

The hydrolysis of the $OC{l}^{-}$ ion
$OC{l}^{-}+H_{2}O\to HOCl+O{H}^{-}$

The hydrolysis constant of the $OC{l}^{-}$ ion
$K_h=\frac{\left[HOCl\right]\left[O{H}^{-}\right]}{\left[OC{l}^{-}\right]}$

$K_a\times K_h=K_w=\left[H_3{O}^{+}\right]\left[O{H}^{-}\right]={10}^{-14}$

$3.0\times {10}^{-8}\times K_h={10}^{-14}$

$K_h=3.33\times {10}^{-7}$