Question

The dissociation constant of $ HA $ is $ 2 \times {10^{ - 8}} $ . The hydrolysis constant of $ NaA $ is
A. $ 1. $ $ 2 \times {10^{ - 8}} $
B. $ 2. $ $ 4 \times {10^{ - 5}} $
C. $ 3. $ $ 5 \times {10^{ - 7}} $
D. $ 4. $ $ 7 \times {10^{ - 3}} $

Answer 1

Hint: A given salt on hydrolysis water produces three types of solution (Acidic, Neutral and Alkaline) depending on its ion. This is due to the fact that contained ions can react with water and thereby produce an acidic or an alkaline solution. For according to the reaction
Hydrolysis of Anion,
   $ {A^ - } + {H_2}O \to HA + O{H^ - } $
Hydrolysis of Cation,
   $ {B^ + } + 2{H_2}O \to BOH + {H_3}{O^ + } $

Compete answer:
In our question it is given that, The value dissociation constant $ \left( {{K_a}} \right) $ of $ HA $ is $ 2 \times {10^{ - 8}}. $ We have to find the value of hydrolysis constant $ \left( {{K_h}} \right) $ of $ NaA $ , to find the hydrolysis constant we must have relation between hydrolysis constant and dissociation constant
Now, Reaction of hydrolysis of $ HA $ ,
   $ HA + {H_2}{O^{}} \rightleftharpoons {H_3}{O^ + } + {A^ - } $
Then, $ {K_a} = \dfrac{{\left[ {{H_3}{O^ + }} \right]\left[ {{A^ - }} \right]}}{{\left[ {HA} \right]}} $
And, Reaction of hydrolysis $ NaA $ is,
   $ {A^ - } + {H_2}O \rightleftharpoons HA + O{H^ - } $
Then, \[{K_h} = \dfrac{{\left[ {HA} \right]\left[ {O{H^ - }} \right]}}{{\left[ {{A^ - }} \right]}}\]
Multiply the numerator and denominator by $ \left[ {{H_3}{O^ + }} \right] $ , we have the above equation
 \[{K_h} = \dfrac{{\left[ {HA} \right]\left[ {O{H^ - }} \right]}}{{\left[ {{A^ - }} \right]}}\dfrac{{\left[ {{H_3}{O^ + }} \right]}}{{\left[ {{H_3}{O^ + }} \right]}}\] \[\because \left[ {\dfrac{1}{{{K_a}}} = \dfrac{{\left[ {{A^ - }} \right]\left[ {{H_3}{O^ + }} \right]}}{{\left[ {HA} \right]}}} \right]\]
By rearranging we get
   $ {K_h} = \dfrac{{{K_w}}}{{{K_a}}} $
Where $ {K_h} $ is hydrolysis constant, $ {K_a} $ is dissociation constant and $ {K_w} $ is water dissociation constant its value is $ {K_w} = 1 \times {10^{ - 14}} $ .
Now solving the question using above relation we get
 \[{K_h} = \dfrac{{1 \times {{10}^{ - 14}}}}{{2 \times {{10}^{ - 8}}}}\]
   $ = 5 \times {10^{ - 7}} $
Thus we get the value of the hydrolysis constant of $ NaA $ is $ 5 \times {10^{ - 7}} $ .

So the correct option is $ \left( 3 \right) $ .

Additional information:
pure water undergoes a reversible reaction in which both $ {H^ + } $ and $ O{H^ - } $ is generated. The equilibrium constant for this reaction is called the water dissociation constant. It is denoted by \[{K_w}\] .

Note:
When a salt is dissolved in water, the nature of the resulting salt depends upon the extent to which either one or both the ions interact with water. Neither of the two ions of a salt formed from a strong acid and strong base undergoes hydrolysis. Therefore when a salt is dissolved in water, its $ pH $ remains the same. The solution is neutral.