Answer
394.2k+ views
Hint:
Degree of Hydrolysis could be defined as the fraction (or percentage) of the total salt which is hydrolyzed at equilibrium.It is generally represented by ‘h’.
\[h=\dfrac{Number\text{ }of\text{ }moles\text{ }of\text{ }salt\text{ }hydrolysed}{Total\text{ }number\text{ }of\text{ }moles\text{ }of\text{ }salt\text{ }taken}\]
Also,
$h=\sqrt{{{k}_{h}}}$
Where ${{k}_{h}}=$ hydrolysis constant
For a given reaction:
\[{{A}^{-}}+{{H}_{2}}O\rightleftharpoons HA+O{{H}^{-}}\]
${{k}_{h}}=\dfrac{[HA][O{{H}^{-}}]}{[{{A}^{-}}][{{H}_{2}}O]}$
Since the concentration of ${{H}_{2}}O$ is very large, it is practically considered as constant. Therefore,
${{k}_{h}}=\dfrac{[HA][O{{H}^{-}}]}{[{{A}^{-}}]}$
If the acidity constant$({{k}_{a}})$ and basicity constant $({{k}_{b}})$ are provided then,
${{k}_{h}}=\dfrac{{{k}_{w}}}{{{k}_{a}}\times {{k}_{b}}}$
Complete Solution :
The equation of hydrolysis of$C{{H}_{3}}COON{{H}_{4}}$is as follows:
$C{{H}_{3}}COON{{H}_{4}}+{{H}_{2}}O\rightleftharpoons C{{H}_{3}}COOH+N{{H}_{4}}OH$
$N{{H}_{4}}OH\rightleftarrows N{{H}_{3}}+{{H}_{2}}O$
From the given reactions, it can be observed that $C{{H}_{3}}COON{{H}_{4}}$ is obtained from a weak acid $(C{{H}_{3}}COOH)$ and a weak base$(N{{H}_{4}}OH)$ .
Since the degree of hydrolysis of a salt of weak acid and weak base is given by:
$h=\sqrt{{{k}_{h}}}$ (For h very small)
$h=\sqrt{\dfrac{{{k}_{w}}}{{{k}_{a}}\times {{k}_{b}}}}$
Therefore, the degree of hydrolysis of $C{{H}_{3}}COON{{H}_{4}}$ is independent of its concentration.
Note: Salts of strong acids and strong bases do not undergo hydrolysis (they undergo only ionization) hence the resulting aqueous solution is neutral.
- For salts of strong acids and weak bases, the degree of hydrolysis is given by:
$h=\sqrt{\dfrac{{{K}_{h}}}{c}}$
- For salts of strong bases and weak acids, the degree of hydrolysis is given by:
$h=\sqrt{\dfrac{{{K}_{h}}}{{{k}_{a}}\times c}}$
Hence for both these types of salts, their degree of hydrolysis is not independent of the concentration .
Degree of Hydrolysis could be defined as the fraction (or percentage) of the total salt which is hydrolyzed at equilibrium.It is generally represented by ‘h’.
\[h=\dfrac{Number\text{ }of\text{ }moles\text{ }of\text{ }salt\text{ }hydrolysed}{Total\text{ }number\text{ }of\text{ }moles\text{ }of\text{ }salt\text{ }taken}\]
Also,
$h=\sqrt{{{k}_{h}}}$
Where ${{k}_{h}}=$ hydrolysis constant
For a given reaction:
\[{{A}^{-}}+{{H}_{2}}O\rightleftharpoons HA+O{{H}^{-}}\]
${{k}_{h}}=\dfrac{[HA][O{{H}^{-}}]}{[{{A}^{-}}][{{H}_{2}}O]}$
Since the concentration of ${{H}_{2}}O$ is very large, it is practically considered as constant. Therefore,
${{k}_{h}}=\dfrac{[HA][O{{H}^{-}}]}{[{{A}^{-}}]}$
If the acidity constant$({{k}_{a}})$ and basicity constant $({{k}_{b}})$ are provided then,
${{k}_{h}}=\dfrac{{{k}_{w}}}{{{k}_{a}}\times {{k}_{b}}}$
Complete Solution :
The equation of hydrolysis of$C{{H}_{3}}COON{{H}_{4}}$is as follows:
$C{{H}_{3}}COON{{H}_{4}}+{{H}_{2}}O\rightleftharpoons C{{H}_{3}}COOH+N{{H}_{4}}OH$
$N{{H}_{4}}OH\rightleftarrows N{{H}_{3}}+{{H}_{2}}O$
From the given reactions, it can be observed that $C{{H}_{3}}COON{{H}_{4}}$ is obtained from a weak acid $(C{{H}_{3}}COOH)$ and a weak base$(N{{H}_{4}}OH)$ .
Since the degree of hydrolysis of a salt of weak acid and weak base is given by:
$h=\sqrt{{{k}_{h}}}$ (For h very small)
$h=\sqrt{\dfrac{{{k}_{w}}}{{{k}_{a}}\times {{k}_{b}}}}$
Therefore, the degree of hydrolysis of $C{{H}_{3}}COON{{H}_{4}}$ is independent of its concentration.
Note: Salts of strong acids and strong bases do not undergo hydrolysis (they undergo only ionization) hence the resulting aqueous solution is neutral.
- For salts of strong acids and weak bases, the degree of hydrolysis is given by:
$h=\sqrt{\dfrac{{{K}_{h}}}{c}}$
- For salts of strong bases and weak acids, the degree of hydrolysis is given by:
$h=\sqrt{\dfrac{{{K}_{h}}}{{{k}_{a}}\times c}}$
Hence for both these types of salts, their degree of hydrolysis is not independent of the concentration .
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