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# What will be the pH and percent degree of hydrolysis respectively for the salt BA of 0.1 M concentration?Given: ${{K}_{a}}\left( HA \right)={{10}^{-6}}$and ${{K}_{b}}\left( BOH \right)={{10}^{-6}}$(A) 5, 1%(B)7, 10%(C) 9, 0.01%(D) 7, 0.01%

Last updated date: 23rd Jul 2024
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Hint:< For a salt of weak acid and weak base the pH of the solution can be calculated by the following formula-
$pH=7+\dfrac{1}{2}\left( p{{K}_{a}}-p{{K}_{b}} \right)$
For the same salt, the degree of hydrolysis can be calculated using the following formula:
$\alpha =\sqrt{\dfrac{{{K}_{w}}}{{{K}_{a}}{{K}_{b}}}}$

Complete step by step solution:
First of all, we have to identify the salt. We can do that by analysing the question given to us. Here, we have been given the equilibrium constant of both the acid and the base that produce the salt. That means, both of them are weak because if they were strong, then they would show complete dissociation in the solution. So, the salt given to us is that of a weak acid and weak base. The reaction of these two to produce the salt is given below-
$BOH+HA\rightleftharpoons BA+{{H}_{2}}O$
For this type of salt, it’s pH is given by the following formula:
$pH=7+\dfrac{1}{2}\left( p{{K}_{a}}-p{{K}_{b}} \right)$
As you already know, that $p{{K}_{a}}$and $p{{K}_{b}}$are the negative logarithm of ${{K}_{a}}$and ${{K}_{b}}$respectively. As we have already been provided with the latter values, we can easily calculate the former ones. We have done it as shown below:
$\Rightarrow p{{K}_{a}}=-\log \left( {{K}_{a}} \right)=-\log \left( {{10}^{-6}} \right)=6=p{{K}_{b}}$
As the values of ${{K}_{a}}$and ${{K}_{b}}$are the same, so the $p{{K}_{a}}$and $p{{K}_{b}}$are also same. Putting them in the above equation we get,
$\Rightarrow pH=7+\dfrac{1}{2}\left( p{{K}_{a}}-p{{K}_{b}} \right)=7+\dfrac{1}{2}\left( 6-6 \right)$
$\Rightarrow pH=7$

So, the pH of the solution is 7, in other words, the solution is neutral.
For solutions where the dilution is pretty high, we can use the following formula to calculate the degree of hydrolysis ($\alpha$) of the salt in solution-
$\Rightarrow \alpha =\sqrt{\dfrac{{{K}_{w}}}{{{K}_{a}}{{K}_{b}}}}$
Where, ${{K}_{w}}$is the ionic product of water and has the constant value of ${{10}^{-14}}$; ${{K}_{a}}$and ${{K}_{b}}$are the equilibrium constants of the dissociation of weak acids and bases, the values of which are already given in the question. So, putting them in the above equation we get-
$\Rightarrow \alpha = \sqrt{\dfrac{{{K}_{w}}}{{{K}_{a}}{{K}_{b}}}}=\sqrt{\dfrac{{{10}^{-14}}}{{{10}^{-6}}\times {{10}^{-6}}}}$
$\Rightarrow \alpha =0.1$

The percent degree of hydrolysis is 10%.

Therefore, the correct option is (B) 7, 10%.

Note: The formula used to calculate the degree of hydrolysis is only valid for weak acid and weak base. It should also be noted that the outcome here is to be multiplied with 100 for getting the percent degree of hydrolysis. The degree of hydrolysis for a salt of a weak acid and a weak base is independent of the concentration of the solution.