Answer
Verified
399k+ views
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.
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.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main
How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE
Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE
Trending doubts
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Write an application to the principal requesting five class 10 english CBSE
Difference Between Plant Cell and Animal Cell
a Tabulate the differences in the characteristics of class 12 chemistry CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
Discuss what these phrases mean to you A a yellow wood class 9 english CBSE
List some examples of Rabi and Kharif crops class 8 biology CBSE