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Question: The dissociation constant of \( HA \) is \( 2 \times {10^{ - 8}} \) . The hydrolysis constant of \( ...

The dissociation constant of HAHA is 2×10−82 \times {10^{ - 8}} . The hydrolysis constant of NaANaA is
A. 1.1. 2×10−82 \times {10^{ - 8}}
B. 2.2. 4×10−54 \times {10^{ - 5}}
C. 3.3. 5×10−75 \times {10^{ - 7}}
D. 4.4. 7×10−37 \times {10^{ - 3}}

Explanation

Solution

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−+H2O→HA+OH−{A^ - } + {H_2}O \to HA + O{H^ - }
Hydrolysis of Cation,
B++2H2O→BOH+H3O+{B^ + } + 2{H_2}O \to BOH + {H_3}{O^ + }

Compete answer:
In our question it is given that, The value dissociation constant (Ka)\left( {{K_a}} \right) of HAHA is 2×10−8.2 \times {10^{ - 8}}. We have to find the value of hydrolysis constant (Kh)\left( {{K_h}} \right) of NaANaA , to find the hydrolysis constant we must have relation between hydrolysis constant and dissociation constant
Now, Reaction of hydrolysis of HAHA ,
HA+H2O⇌H3O++A−HA + {H_2}{O^{}} \rightleftharpoons {H_3}{O^ + } + {A^ - }
Then, Ka=[H3O+][A−][HA]{K_a} = \dfrac{{\left[ {{H_3}{O^ + }} \right]\left[ {{A^ - }} \right]}}{{\left[ {HA} \right]}}
And, Reaction of hydrolysis NaANaA is,
A−+H2O⇌HA+OH−{A^ - } + {H_2}O \rightleftharpoons HA + O{H^ - }
Then, Kh=[HA][OH−][A−]{K_h} = \dfrac{{\left[ {HA} \right]\left[ {O{H^ - }} \right]}}{{\left[ {{A^ - }} \right]}}
Multiply the numerator and denominator by [H3O+]\left[ {{H_3}{O^ + }} \right] , we have the above equation
Kh=[HA][OH−][A−][H3O+][H3O+]{K_h} = \dfrac{{\left[ {HA} \right]\left[ {O{H^ - }} \right]}}{{\left[ {{A^ - }} \right]}}\dfrac{{\left[ {{H_3}{O^ + }} \right]}}{{\left[ {{H_3}{O^ + }} \right]}} ∵[1Ka=[A−][H3O+][HA]]\because \left[ {\dfrac{1}{{{K_a}}} = \dfrac{{\left[ {{A^ - }} \right]\left[ {{H_3}{O^ + }} \right]}}{{\left[ {HA} \right]}}} \right]
By rearranging we get
Kh=KwKa{K_h} = \dfrac{{{K_w}}}{{{K_a}}}
Where Kh{K_h} is hydrolysis constant, Ka{K_a} is dissociation constant and Kw{K_w} is water dissociation constant its value is Kw=1×10−14{K_w} = 1 \times {10^{ - 14}} .
Now solving the question using above relation we get
Kh=1×10−142×10−8{K_h} = \dfrac{{1 \times {{10}^{ - 14}}}}{{2 \times {{10}^{ - 8}}}}
=5×10−7= 5 \times {10^{ - 7}}
Thus we get the value of the hydrolysis constant of NaANaA is 5×10−75 \times {10^{ - 7}} .

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

Additional information:
pure water undergoes a reversible reaction in which both H+{H^ + } and OH−O{H^ - } is generated. The equilibrium constant for this reaction is called the water dissociation constant. It is denoted by Kw{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 pHpH remains the same. The solution is neutral.