Question

Relationship between hydrolysis constant and dissociation constant are given. which is the correct formula to $MgCl_2$ ?
A. ${K_h} = \dfrac{{{K_w}}}{{{K_a}}}$
B. ${K_h} = \dfrac{{{K_w}}}{{{K_b}}}$
C. ${K_h} = \dfrac{{{K_w}}}{{{K_a} \times {K_b}}}$
D. ${K_w} = \dfrac{{{K_h}}}{{{K_b}}}$

Answer 1

Hint: We need to know the equilibrium constant for hydrolysis reaction called hydrolysis constant. It is denoted by $K_h$ . An acid dissociation constant ( $K_a$ ) is a quantitative measure of the strength of an acid in solution. A base dissociation constant( $K_b$ ) is a quantitative measure of the strength of the base in solution.

Complete step by step answer:
Now let us give the hydrolytic reaction of Magnesium chloride
First let us ionize the compound,
$MgC{l_2} \rightleftarrows M{g^ + } + 2C{l^ - }$
Now the hydrolytic reaction of ionized product of $MgCl_2$ is
$M{g^{2 + }} + 2C{l^ - } + 2{H_2}O \rightleftharpoons Mg{(OH)_2} + 2HCl$
We know that $MgCl_2$ is a salt made up of weak base Mg(OH)2 and strong acid HCl. Since Mg(oH)2 is a weak base it hydrolysis easily in water.
Now let us write the hydrolysis of Mg(OH)2
\[M{g^{2 + }} + 2{H_2}O \rightleftharpoons Mg{(OH)_2} + 2{H^ + }\]
[H2O] = 1
Now the hydrolysis constant can be written as
${K_h} = \dfrac{{\left[ {Mg{{(OH)}_2}} \right]\left[ {{H^ + }} \right]}}{{\left[ {M{g^{2 + }}} \right]}}$
Now let us calculate KW
${H_2}O \rightleftarrows {H^ + } + O{H^ - }$
${K_W} = \left[ {{H^ + }} \right]\left[ {O{H^ - }} \right]$ …...(1)
Now let us calculate equilibrium constant, since we are considering weak base it is denoted as $K_b$
Now let us write the dissociation equation,
$Mg{\left( {OH} \right)_2} \rightleftarrows M{g^ + } + 2O{H^ - }$
Now $K_b$ can be given as,
Dissociation constant
$\left( {{K_b}} \right){\text{ = }}\dfrac{{\left[ {M{g^ + }} \right]{{\left[ {O{H^ - }} \right]}^2}}}{{\left[ {Mg{{\left( {OH} \right)}_2}} \right]}}$………(2)
Divide equation 1 by 2
\[\dfrac{{{K_w}}}{{{K_b}}} = \dfrac{{\dfrac{{\left[ {{H^ + }} \right]\left[ {O{H^ - }} \right]}}{1}}}{{\dfrac{{\left[ {M{g^ + }} \right]\left[ {O{H^ - }} \right]}}{{\left[ {Mg{{\left( {OH} \right)}_2}} \right]}}}}\]
As [OH-] is present in both numerator and denominator it is cancelled out, then the equation becomes
\[\dfrac{{{K_w}}}{{{K_b}}} = \dfrac{{\left[ {{H^ + }} \right]\left[ {Mg{{\left( {OH} \right)}_2}} \right]}}{{\left[ {M{g^ + }} \right]}} = {K_h}\]
Therefore ${K_h} = \dfrac{{{K_w}}}{{{K_b}}}$

Therefore, the option B is correct.

Note: We have to remember that the hydrolysis constant and dissociation constant are inversely proportional.
In general it can be given as
First step is to write the hydrolysis equilibrium of the compound. For example
$A + {H_2}O \rightleftarrows HA + OH$
Now the hydrolysis constant for the above equation is ,
${K_h} = \dfrac{{\left[ {HA} \right]\left[ {OH} \right]}}{{\left[ A \right]}}$ $\because \left[ {{H_2}O} \right] = 1$ (1)
In the above equation HA is a weak acid, then the equilibrium constant is represented as $K_a$ , if the compound has a weak base then the equilibrium constant is represented as $K_b$ .
Now the dissociation of weak acid HA is
$HA \rightleftarrows {H^ + } + {A^ - }$
The equilibrium constant of the weak acid is given as
${K_a} = \dfrac{{\left[ {{H^ + }} \right]\left[ {{A^ - }} \right]}}{{\left[ {HA} \right]}}$ (2)
Now calculate Kw, it is the equilibrium constant of water
${K_W} = \left[ {{H^ + }} \right]\left[ {O{H^ - }} \right]$ (3)
Divide equation (3) by(2)
$\dfrac{{{K_w}}}{{{K_a}}} = \dfrac{{\dfrac{{\left[ {{H^ + }} \right]\left[ {O{H^ - }} \right]}}{1}}}{{\dfrac{{\left[ {{H^ + }} \right]\left[ {{A^ - }} \right]}}{{\left[ A \right]}}}}$
As [H+] is present in both numerator and denominator it is cancelled out, then the equation becomes
$\dfrac{{{K_w}}}{{{K_a}}} = \dfrac{{\left[ {HA} \right]\left[ {O{H^ - }} \right]}}{{\left[ A \right]}} = {K_h}$
Therefore, for weak acid the hydrolysis constant is
$\dfrac{{{K_w}}}{{{K_a}}} = {K_h}$ .