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# How much must a $\text{0}\text{.2 M }$ solution of sodium acetate be diluted at $\text{ 25 }\!\!{}^\circ\!\!\text{ C }$ in order to double the degree of hydrolysis?

Last updated date: 19th Sep 2024
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Hint: The salt of weak acid or base undergoes the hydrolysis reaction. The general hydrolysis reaction of the salt is given as follows:
$\text{ }{{\begin{matrix} {{\text{B}}^{-}} & \text{+} & {{\text{H}}_{\text{2}}}\text{O} & \rightleftharpoons & \text{BH} & \text{+} & \text{OH} \\ \end{matrix}}^{-}}\text{ }$
The degree of hydrolysis for a base is given as,
$\text{ }{{\text{K}}_{\text{h}}}\text{ = }\dfrac{\left[ \text{BH} \right]\left[ \text{O}{{\text{H}}^{\text{-}}} \right]}{\left[ {{\text{B}}^{\text{-}}} \right]}\text{ = }{{\text{h}}^{\text{2}}}\text{.C }$
Where ‘h’ is the degree of hydrolysis and C is the concentration of the solution.

The hydrolysis reaction is the addition of the water molecule to molecules, such that the water molecules rupture or more bond and form corresponding products.
The degree of hydrolysis is the fraction of the total amount of the salt which is hydrolysed at the equilibrium. The sodium acetate undergoes the hydrolysis reaction. The reaction is as follows:
$\text{ }\begin{matrix} \text{C}{{\text{H}}_{\text{3}}}\text{COONa} & \text{+} & {{\text{H}}_{\text{2}}}\text{O} & \rightleftharpoons & \text{C}{{\text{H}}_{\text{3}}}\text{COOH} & \text{+} & \text{NaOH} \\ \end{matrix}$

We have given the following data:
The concentration of sodium acetate $\text{ (C}{{\text{H}}_{\text{3}}}\text{COONa) }$ solution is$\text{ 0}\text{.2 M }$.
We have to find the volume of the sodium acetate solution when the degree of hydrolysis is doubled.
The degree of hydrolysis for salt is given as:
$\text{ }{{\text{K}}_{\text{h}}}\text{ = }{{\text{h}}^{2}}\text{ C }$
Where $\text{ }{{\text{K}}_{\text{h}}}\text{ }$ is the equilibrium constant for the hydrolysis reaction
‘h’ is the degree of hydrolysis of the reaction
‘C’ is the concentration of the solution.

Let consider the concentration of sodium acetate as $\text{ }{{\text{C}}_{\text{1}}}\text{ }$ then the equilibrium constant for the sodium acetate is written as:
$\text{ }{{\text{K}}_{\text{h}}}\text{ = }{{\text{h}}^{2}}\text{ }{{\text{C}}_{\text{1}}}\text{ }$
Here, $\text{ }{{\text{C}}_{\text{1}}}\text{ = 0}\text{.2 M }$ and consider ‘h’ as the degree of hydrolysis. Then, we have
$\text{ }{{\text{K}}_{\text{h}}}\text{ = (h}{{\text{)}}^{2}}\text{ }\left( 0.2 \right)\text{ }$ (1)
When the degree of hydrolysis is doubled, the concentration would be,
$\text{ }{{\text{K}}_{\text{h}}}\text{ = (2h}{{\text{)}}^{2}}\text{ }{{\text{C}}_{2}}\text{ }$ (2)
Divide the equation (1) by equation (2). We get,
$\text{ }\begin{matrix} \dfrac{{{\text{K}}_{\text{h}}}}{{{\text{K}}_{\text{h}}}} & = & \dfrac{{{\text{(h)}}^{2}}\text{ }\left( 0.2 \right)}{{{\text{(2h)}}^{2}}\text{ }{{\text{C}}_{2}}} \\ 1 & = & \dfrac{0.2}{4{{\text{C}}_{2}}\text{ }} \\ {{\text{C}}_{2}}\text{ } & = & 0.05\text{ M} \\ \end{matrix}\text{ }$
The concentration of solution when we double the degree of hydrolysis is equal to $\text{ }0.05\text{ M }$ .
We are interested to find the volume of the solution. We apply the following formula,$\text{ }\begin{matrix} {{\text{M}}_{\text{1}}}{{\text{V}}_{\text{1}}} & = & {{\text{M}}_{\text{2}}}{{\text{V}}_{\text{2}}} \\ \left( \text{0}\text{.2} \right)\left( {{\text{V}}_{\text{1}}} \right) & = & \left( \text{0}\text{.05} \right){{\text{V}}_{\text{2}}} \\ {{\text{V}}_{\text{2}}} & = & \dfrac{\text{0}\text{.2}}{\text{0}\text{.05}}{{\text{V}}_{\text{1}}} \\ \therefore {{\text{V}}_{\text{2}}} & = & 4\text{ }{{\text{V}}_{\text{1}}} \\ \end{matrix}$
Thus, the volume of the $\text{ 0}\text{.2 M }$ sodium acetate solution should be four times the initial volume so that the degree of hydrolysis would be double than the initial degree of hydrolysis.

Note: Note that the salt of the strong base or the acid does not undergo the hydrolysis reaction. The hydrolysis reaction is the addition of water molecules to the salt. The salt of strong acid or base undergoes ionization only thus always resulting in the neutral aqueous solution. The hydrolysis of salt of weak base results in the hydroxide ion, thus the solution is basic.