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If the ${K}_{b}$ value in the hydrolysis reaction, ${ B }^{ + }\quad +\quad { H }_{ 2 }O\quad \longrightarrow \quad BOH\quad +\quad { H }^{ + }$ is $1.0 \times {10}^{-6}$, then the hydrolysis constant of the salt would be:
A. $1.0 \times {10}^{-6}$
B. $1.0 \times {10}^{-7}$
C. $1.0 \times {10}^{-8}$
D. $1.0 \times {10}^{-9}$

Last updated date: 13th Jun 2024
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Hint: The base dissociation constant is a measure of the extent to which a base dissociates into its component ions in water. Hydrolysis constant can be defined as the equilibrium constant for a hydrolysis reaction. The base dissociation is inversely proportional to the hydrolysis constant.

Complete step by step answer:
> ${K}_{b}$ is the base dissociation constant of the reaction. It tells us how completely the base dissociates into its component ions in water. The larger the value of ${K}_{b}$ indicates a high level of dissociation of a strong base.
> Pure water also undergoes auto-ionization to form ${H}_{3}{O}^{+}$ and ${OH}^{-}$ ions and this reaction in which auto-ionization takes place always stays in equilibrium. Therefore, the equilibrium constant for auto-ionization of water is known as ${K}_{w}$ and its value is constant at a particular temperature. At room temperature its value is $1.0 \times {10}^{-14}$.
> Hydrolysis constant is the equilibrium constant of a hydrolysis reaction and it is denoted by ${K}_{H}$. The hydrolysis constant is related to the ionic product of water, ${K}_{w}$ and the base dissociation, ${K}_{b}$.
This relation is given as follows.

${ K }_{ H }\quad =\quad \dfrac { { K }_{ w } }{ { K }_{ b } }$

Now, for the given hydrolysis reaction:

${ B }^{ + }\quad +\quad { H }_{ 2 }O\quad \longrightarrow \quad BOH\quad +\quad { H }^{ + }$

Now, the expression for hydrolysis constant is:

${ K }_{ H }\quad =\quad \dfrac { { K }_{ w } }{ { K }_{ b } }$ ------(1)

And is also given that ${K}_{b} = 1.0 \times {1.0}^{-6}$ and we know that the value of ${K}_{w}$ is $1.0 \times {10}^{-14}$. Substituting, these values in equation (1), we get

${ K }_{ H }\quad =\quad \dfrac { 1.0\quad \times \quad { 10 }^{ -14 } }{ 1.0\quad \times \quad { 10 }^{ -6 } }$

${ K }_{ H }\quad =\quad 1.0\quad \times \quad { 10 }^{ -8 }$
Therefore, the value of hydrolysis constant is $1.0 \times {10}^{-8}$.

Hence, the correct answer is option (C).

Note: The auto-ionization reaction always stays in equilibrium because the component ions after the dissociation of water are hydronium ion and hydroxide ion. And the hydronium ion is a very strong acid and the hydroxide ion is a very strong base. Thus they associate again to form water molecules. So, the water molecules and ions always stay in equilibrium.