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

Hint: The hydrolysis constant can be determined using the formulae,
\[{{K}_{a}}\times {{K}_{h}}={{K}_{w}}=\left[ {{H}_{3}}{{O}^{+}} \right]\left[ O{{H}^{-}} \right]={{10}^{-14}}\]
Where \[{{K}_{a}}\], is the acid dissociation constant
\[{{K}_{h}}\], is the hydrolysis constant
\[{{K}_{\begin{smallmatrix}
 w \\

\end{smallmatrix}}}\], is the dissociation constant for ionization constant of water

Complete Solution :
Let’s check the data given in the reaction.
It is given that the ionization constant of HOCl is $3.0\times {{10}^{-8}}$

Now let’s write the equation:
\[HOCl+{{H}_{2}}O\rightleftarrows {{H}_{3}}{{O}^{+}}+OCl\]
From the equation let’s write the equation for${{K}_{a}}$,
\[{{K}_{a}}=\dfrac{\left[ {{H}_{3}}{{O}^{+}} \right]\left[ OC{{l}^{-}} \right]}{\left[ HOCl \right]}=3.0\times {{10}^{-8}}\]

And now let’s consider the hydrolysis of the $OC{{l}^{-}}$ ion,
\[OC{{l}^{-}}+{{H}_{2}}O\to HOCl+O{{H}^{-}}\]
From this equation we can write the hydrolysis constant of the\[OC{{l}^{-}}\] ion as
 \[{{K}_{h}}=\dfrac{\left[ HOC{{l}^{{}}} \right]\left[ O{{H}^{-}} \right]}{\left[ OC{{l}^{-}} \right]}\]

And we know the relation:
\[{{K}_{a}}\times {{K}_{h}}={{K}_{w}}=\left[ {{H}_{3}}{{O}^{+}} \right]\left[ O{{H}^{-}} \right]={{10}^{-14}}\]
By substituting the values we will get ${{K}_{h}}$,
\[{{K}_{h}}=\dfrac{{{10}^{-14}}}{3.0\times {{10}^{-8}}}\]
\[{{K}_{h}}=3.33\times {{10}^{-7}}\]
So, the correct answer is “Option B”.


Note: It is always better to study the standard equations and change the values according to the need of the question. Writing the equation of equilibrium should be given more concern as we derive further equations and steps from the standard equation of equilibrium. Addition and subtraction of powers should be given concern while changing denominator and numerator.