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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?

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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%

Additional Information:
Ionisation constant corresponds to the strength of an acid or a base.
-In an equilibrium reaction the backward and the forward reaction rates are the same. But the concentration varies. Since the concentration gives us the idea of how much the substance has dissociated, we consider the ratios of concentrations of reactant and products to calculate ionisation constant K. In an acid base equilibrium reaction, it is equilibrium constant which gives the extent of dissociation. For strong acids and strong bases, the ionisation constant is easy to calculate. But for weak acid and weak bases, their reaction is usually in equilibrium, so its calculation is complex.

Note: Ionisation and dissociation are not the same. Dissociation is the process where the charged particles already existing in the compound get separated. Ionisation is the process where new charged particles are formed which were absent in the previous compound.

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