Question

# The ionisation constant dimethylamine is $5.4\times {{10}^{-4}}$. Calculate its degree of ionisation in its 0.02M solution. What percentage of dimethylamine is ionised if the solution is also 0.1M NaOH?

Hint: Ionisation constant is the constant which depends on the equilibrium between the ions and the non-ionised molecules in a solution. The degree of ionisation is the ratio between the ionised and the neutral molecules in the solution.

Complete step by step solution:
-We know that the equation for calculating degree of ionisation is given as
$y=\sqrt{\dfrac{{{K}_{b}}}{c}}$, here y is degree of ionisation and ionisation constant is ${{K}_{b}}$=$5.4\times {{10}^{-4}}$
${{K}_{b}}$=ionisation constant= $5.4\times {{10}^{-4}}$and concentration is c=0.02M=$2\times {{10}^{-2}}$M. These values are given in the question.
Now let us substitute the values in the above equation, so we get
$y=\sqrt{\dfrac{5.4\times {{10}^{-4}}}{2\times {{10}^{-2}}}}$
=0.164

Now let’s solve the next part of the question.
Let us consider, x be the amount of dimethylamine which dissociate in the presence of 0.1M NaOH
We can write the equation as $\underset{0.02-x}{\mathop{{{(C{{H}_{3}})}_{2}}NH}}\,+{{H}_{2}}O\rightleftharpoons \underset{x}{\mathop{{{(C{{H}_{3}})}_{2}}NH_{2}^{+}}}\,+\underset{x}{\mathop{O{{H}^{-}}}}\,$
We know that NaOH undergoes complete ionisation, so its equation can be written as
$NaOH\rightleftharpoons \underset{0.1M}{\mathop{N{{a}^{+}}}}\,+\underset{0.1M}{\mathop{O{{H}^{-}}}}\,$
The equilibrium concentrations are given as:
$[{{(C{{H}_{3}})}_{2}}NH]=0.02-x\approx 0.02$
$[{{(C{{H}_{3}})}_{2}}NH_{2}^{+}]=x$
$[O{{H}^{-}}]=0.1+x\approx 0.1$

We know the expression for equilibrium constant is the product of the concentration of the products divided by the product of concentration of the reactants. In this case it is written as( and substituting the values of the concentrations)
${{K}_{b}}=\frac{[{{(C{{H}_{3}})}_{2}}NH_{2}^{+}][O{{H}^{-}}]}{[{{(C{{H}_{3}})}_{2}}NH]}=\dfrac{x\times 0.1}{0.02}$
Solving this equation, we will get the value of x.
$x=1.08\times {{10}^{-4}}$
Thus, the amount of dimethylamine dissociated in the presence of 0.1M NaOH is
$x=1.08\times {{10}^{-4}}=[{{(C{{H}_{3}})}_{2}}NH_{2}^{+}]$
For calculating the percentage of dimethylamine dissociation, it is given by the equation
$\dfrac{1.08\times {{10}^{-4}}}{0.02}\times 100$, where 0.02 is the initial concentration of dimethylamine. We get,
$=0.54%$
Therefore, the percentage of dimethylamine is 0.54%