
In a hydrolysis of a salt of weak acid and strong base ${{A}^{-}}+{{H}_{2}}O\rightleftharpoons HA+O{{H}^{-}}$, the hydrolysis constant $({{K}_{h}})$is equal to:
A. $\dfrac{{{K}_{w}}}{{{K}_{a}}}$
B. $\dfrac{{{K}_{w}}}{{{K}_{b}}}$
C. $\sqrt{\dfrac{{{K}_{a}}}{C}}$
D. $\dfrac{{{K}_{w}}}{{{K}_{a}}\times {{K}_{b}}}$
Answer
163.2k+ views
Hint: Hydrolysis is the chemical breakdown of the substances by water. A solution of weak acid mix with a solution of strong base to produce the conjugate base of weak acid and the conjugate acid of strong base. The conjugate acid of the strong base is a weaker acid than water and has little effect on the acidity of the resulting solution.
Complete Step by Step Answer:
Consider the hydrolysis of weak acid and strong base.
Let us consider the general hydrolysis reaction of a salt of weak acid and strong base which can be written as:
${{A}^{-}}+{{H}_{2}}O\rightleftharpoons HA+O{{H}^{-}}$
This leads to the equilibrium constant expression
${{K}_{h}}=\dfrac{[HA][O{{H}^{-}}]}{[{{A}^{-}}][{{H}_{2}}O]}$
The concentration of water is very large and is regarded as practically constant.
${{K}_{h}}=\dfrac{[HA][O{{H}^{-}}]}{[{{A}^{-}}]}$
And ${{K}_{w}}=[{{H}^{+}}][O{{H}^{-}}]$
For the dissociation of a weak acid HA,
$HA\rightleftharpoons {{H}^{+}}+{{A}^{-}}$
The acid dissociation constant ${{K}_{a}}$is expressed as:
${{K}_{a}}=\dfrac{[{{H}^{+}}][{{A}^{-}}]}{[HA]}$
Then $\dfrac{{{K}_{w}}}{{{K}_{a}}}=\dfrac{[O{{H}^{-}}][HA]}{[{{A}^{-}}]}$=${{K}_{h}}$
Hence $\dfrac{{{K}_{w}}}{{{K}_{a}}}={{K}_{h}}$
Thus, Option (A) is correct.
Note: Solubility product is a type of equilibrium constant whose value depends on the temperature. It is denoted by ${{K}_{sp}}$. It usually increases with the increase in temperature because of the increased solubility. Remember that solubility and the solubility product are different from each other. Solubility of a substance in a solvent is the highest amount of solute that can be dissolved in a solvent whereas the solubility product is an equilibrium constant that gives the equilibrium between the solid solute and its ions that are dissolved in the solution
Complete Step by Step Answer:
Consider the hydrolysis of weak acid and strong base.
Let us consider the general hydrolysis reaction of a salt of weak acid and strong base which can be written as:
${{A}^{-}}+{{H}_{2}}O\rightleftharpoons HA+O{{H}^{-}}$
This leads to the equilibrium constant expression
${{K}_{h}}=\dfrac{[HA][O{{H}^{-}}]}{[{{A}^{-}}][{{H}_{2}}O]}$
The concentration of water is very large and is regarded as practically constant.
${{K}_{h}}=\dfrac{[HA][O{{H}^{-}}]}{[{{A}^{-}}]}$
And ${{K}_{w}}=[{{H}^{+}}][O{{H}^{-}}]$
For the dissociation of a weak acid HA,
$HA\rightleftharpoons {{H}^{+}}+{{A}^{-}}$
The acid dissociation constant ${{K}_{a}}$is expressed as:
${{K}_{a}}=\dfrac{[{{H}^{+}}][{{A}^{-}}]}{[HA]}$
Then $\dfrac{{{K}_{w}}}{{{K}_{a}}}=\dfrac{[O{{H}^{-}}][HA]}{[{{A}^{-}}]}$=${{K}_{h}}$
Hence $\dfrac{{{K}_{w}}}{{{K}_{a}}}={{K}_{h}}$
Thus, Option (A) is correct.
Note: Solubility product is a type of equilibrium constant whose value depends on the temperature. It is denoted by ${{K}_{sp}}$. It usually increases with the increase in temperature because of the increased solubility. Remember that solubility and the solubility product are different from each other. Solubility of a substance in a solvent is the highest amount of solute that can be dissolved in a solvent whereas the solubility product is an equilibrium constant that gives the equilibrium between the solid solute and its ions that are dissolved in the solution
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