   Question Answers

# 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%

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.

View Notes
Specific Heat of Constant Pressure  Ionization Enthalpy and Valency    Spring Constant Formula  What is the Cell Envelope?  What is the Placebo Effect  What is the full form of phd?  NCERT Book Class 11 Chemistry PDF  What is the Difference Between Trade and Commerce?  The Difference Between an Animal that is A Regulator and One that is A Conformer  Important Questions for CBSE Class 11 Chemistry Chapter 11 - The p-Block Elements  Important Questions for CBSE Class 11 Chemistry  Important Questions for CBSE Class 11 Chemistry Chapter 10 - The s-Block Elements  Important Questions for CBSE Class 11 Chemistry Chapter 14 - Environmental Chemistry  Important Questions for CBSE Class 11 Chemistry Chapter 13 - Hydrocarbons  Important Questions for CBSE Class 11 Chemistry Chapter 1 - Some Basic Concepts of Chemistry  Important Questions for CBSE Class 11 Chemistry Chapter 6 - Thermodynamics  Important Questions for CBSE Class 11 Chemistry Chapter 7 - Equilibrium  Important Questions for CBSE Class 11 Chemistry Chapter 9 - Hydrogen  Important Questions for CBSE Class 11 Chemistry Chapter 12 - Organic Chemistry - Some Basic Principles and Techniques  Chemistry Question Paper for CBSE Class 12  CBSE Class 12 Chemistry Question Paper 2020  Chemistry Question Paper for CBSE Class 12 - 2013  Chemistry Question Paper for CBSE Class 12 - 2015  CBSE Class 12 Chemistry Question Paper 2019 - Free PDF  CBSE Class 12 Chemistry Question Paper 2017 - Free PDF  CBSE Class 12 Chemistry Question Paper 2018 - Free PDF  Previous Year Question Paper for CBSE Class 12 Chemistry - 2014  Chemistry Question Paper for CBSE Class 12 - 2016 Set 1 E  Chemistry Question Paper for CBSE Class 12 - 2016 Set 1 S  NCERT Solutions for Class 11 Chemistry Chapter 11 The p-Block Elements In Hindi  NCERT Solutions Class 11 Chemistry  NCERT Solutions for Class 11 Chemistry Chapter 11  Equilibrium NCERT Solutions - Class 11 Chemistry  NCERT Exemplar for Class 11 Chemistry Chapter-11 (Book Solutions)  NCERT Solutions Class 11 Chemistry Chapter 1  NCERT Solutions for Class 11 Chemistry Chapter 10 The S-Block Elements In Hindi  NCERT Solutions for Class 11 Chemistry Chapter 14 Environmental Chemistry In Hindi  NCERT Solutions for Class 11 Chemistry Chapter 9  NCERT Solutions for Class 11 Chemistry Chapter 5  