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# Calculate hydrolysis constant for 0.01 $N$solution of $C{H_3}C{O_2}Na$.$\left( {{{\text{K}}_a}{\text{ }}for{\text{ }}C{H_3}C{O_2}H = 1.810 \times {{10}^{ - 5}}} \right)$

Last updated date: 09th Sep 2024
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Hint: For sodium acetate $\left( {C{H_3}C{O_2}Na} \right)$ molecular weight is equal to equivalent weight.
Hence normality is equal to molarity.
In generalized form, the hydrolysis constant can be described as: ${K_a} = \dfrac{{\left[ {{H_3}{O^ + }} \right] \times [{A^ - }]}}{{\left[ {HA} \right]}}$
Where ${A^ - }$ represents any base, and $HA$ represents any acid.

Step by step solution:
A hydrolysis constant is an equilibrium constant for a hydrolysis reaction.
The equivalent weight of an element or radical is equal to its atomic weight or formula weight divided by the valence it assumes in compounds:
$equivalent{\text{ }}weight = \dfrac{{molecular{\text{ }}weight}}{{valence}}$
The unit of equivalent weight is the atomic mass unit.
For sodium acetate $\left( {C{H_3}C{O_2}Na} \right)$ equivalent weight is equal to molecular weight.
Next form the definition of normality and molarity
$Normality = \dfrac{{equivalent{\text{ }}weight{\text{ }}of{\text{ }}solute}}{{volume{\text{ }}of{\text{ }}solution{\text{ }}in{\text{ }}litre}}$
And $Molarity = \dfrac{{mass{\text{ }}of{\text{ }}solute}}{{volume{\text{ }}of{\text{ }}solution}}$
If the equivalent weight is equal to molecular weight, the normality will also be equal to molarity.
So the concentration can be taken as 0.01 $M$
Hydrolysis reaction of $C{H_3}C{O_2}Na$ can be written as:
$C{H_3}COONa + {H_2}O \to C{H_3}COOH + N{a^ + } + O{H^ - }$
To calculate the $pOH$ of a solution we need to know the concentration of the hydroxide ion in moles per liter (molarity). The $pOH$ is then can be calculated using the expression:
$pOH{\text{ }} = \; - {\text{ }}log{\text{ }}\left[ {O{H^ - }} \right]$
So first we will calculate the value of ${{\text{K}}_b}$ using ${{\text{K}}_a}$ as:
${{\text{K}}_w} = {{\text{K}}_a} \times {{\text{K}}_b}$
On substituting the values
$1 \times {10^{ - 14}} = 1.81 \times {10^{ - 5}} \times {K_b}$
So, ${K_b} = 5.56 \times {10^{ - 10}}$
Next to find out $O{H^ - }$ ions, we will use the formula of ${{\text{K}}_b}$ as:
${K_b} = \dfrac{{\left[ {N{a^ + }} \right]\left[ {O{H^ - }} \right]}}{{C{H_3}COONa}}$
Now let the concentration of $N{a_ + }$ and $O{H^ - }$ be $x$
Now on substituting the values:
$5.56 \times {10^{ - 10}} = \dfrac{{[x] \times [x]}}{{\left( {0.01 - x} \right)}} \\ 5.56 \times {10^{ - 10}}\left( {0.01 - x} \right) = {x^2} \\ {x^2} - 5.56 \times {10^{ - 10}}x + 5.56 \times {10^{ - 10}} \\$
$x = 2.3 \times {10^{ - 6}}M$
Hence $\left[ {N{a^ + }} \right] = \left[ {O{H^ - }} \right] = 2.3 \times {10^{ - 6}}M$
On substituting the value of $\left[ {O{H^ - }} \right]$
$pOH{\text{ }} = \; - {\text{ }}log{\text{ }}\left( {2.3 \times {{10}^{ - 6}}} \right)$
$pOH = 5.64$
Now to find out $pH$we will use the given formula:
$pH = 14 - pOH \\ pH = 14 - 5.6 \\ pH = 8.4 \\$
Note: If $p{K_a}$< $p{K_b}$, $pH$ of the solution will be less than 7 and the solution will be acidic.
If$p{K_a}$= $p{K_b}$, $pH$ of the solution will be equal to 7 and the solution will be neutral.
And if $p{K_a}$> $p{K_b}$ $pH$ of the solution will be more than 7 and the solution will be basic.