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-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%

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.

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