Question

Calculate the degree of hydrolysis of $0.10$solution of $KCN$.Dissociation constant of $KCN = 7.2 \times {10^{ - 10}}$at$25^\circ C$and${K_w} = 1.0 \times {10^{ - 4}}$.

Answer 1

Hint – You can start by describing the dissociation constant, degree of hydrolysis and the ionization constant. Then use the equation${K_h} = \dfrac{{{K_w}}}{{{K_a}}}$to find${K_h}$. Then use the equation$h = \sqrt {\dfrac{{{K_h}}}{c}} $ to find the degree of hydrolysis.

Complete step by step solution:
> Dissociation is a process by which molecules, ionic compounds such as salts or complex compounds get separated into its constituent atom, ions or radicals. The process of dissociation is usually reversed in nature.
> Dissociation Constant $({K_h})$– It is the ratio of concentration of dissociated entities to the concentration of undissociated entities.
For the case of chemical reactions in an equilibrium
$AB\overset {} \leftrightarrows A + B$
${K_d} = \dfrac{{\left[ A \right]\left[ B \right]}}{{\left[ {AB} \right]}}$
Here$\left[ A \right]$,$\left[ B \right]$and$\left[ {AB} \right]$are the concentrations of the corresponding chemical species.
> Degree of hydrolysis $(h)$– It is an equilibrium constant, which represents the degree of a hydrolysis reaction. The highest value of degree of hydrolysis is $1$ and the lowest value is$0$. For example – If we mix salt with water and suppose $80\% $of the salt dissociates in the solution. Then we say the degree of dissociation is 0.80.
Ionization constant of water$({K_w})$ - Water molecules have the tendency to self-ionize. This degree of self-ionization is called Ionization constant of water.
Given, ${K_w} = 1 \times {10^{ - 14}}$
${K_a} = 7.2 \times {10^{ - 10}}$
$c = 0.10M$
We know that the equation of the hydrolysis constant is
${K_h} = \dfrac{{{K_w}}}{{{K_a}}}$
$ \Rightarrow {K_h} = \dfrac{{{{10}^{ - 14}}}}{{7.2 \times {{10}^{ - 10}}}}$
$ \Rightarrow {K_h} = 1.38 \times {10^{ - 5}}$
We also know that the degree of hydrolysis is
$h = \sqrt {\dfrac{{{K_h}}}{c}} $
$ \Rightarrow h = \sqrt {\dfrac{{1.38 \times {{10}^{ - 5}}}}{{0.1}}} $
\[ \Rightarrow h = \sqrt {1.38 \times {{10}^{ - 4}}} \]
\[ \Rightarrow h = 1.18 \times {10^{ - 2}}\]

Note – In such types of problems, we usually assume that the temperature is constant and there is no catalyst involved. But you may encounter problems that have these factors involved. Remember temperature increases dissociation in an endothermic reaction and decreases dissociation in an exothermic reaction.