Question

The dissociation constants of a m-nitrobenzoic acid and acetic acid are $36.0 \times {10^{ - 5}}$ and $1.8 \times {10^{ - 5}}$ respectively. What are their relative strengths?

Answer 1

Hint: In this question to find relative strength we will first find the degree of dissociation of both the given compounds and then to find relative strength we will divide the degree of dissociation of m-nitrobenzoic acid with that of acetic acid.

Complete step by step answer:
As given in question, by dissociation constant we mean the ratio of dissociated ions in product to the original reactants. It is written as ${K_a}$ and by relative strength we mean the ratio of degree of dissociation of the given acids.
To find relative strength, first we need to find the degree of dissociation. By degree of dissociation of a substance we mean the fraction of its molecules dissociating at a given time.
The relation of degree of dissociation with the dissociation constant for weak acids can be given as :
                                                  $\alpha = \sqrt {\dfrac{{{K_a}}}{C}} $ $ - (1)$
 Where, $\alpha = $ Degree of dissociation
                 ${K_a} = $ acid dissociation constant
                $C = $ Concentration of the acidic solution
Now, let the degree of dissociation of m-nitrobenzoic acid be ${\alpha _1}$ and the degree of dissociation of acetic acid be ${\alpha _2}$. Also, as we are calculating the relative strength so we'll take concentration of the acidic solution same in both cases.
Now, given for m-nitrobenzoic acid ${K_a} = 36.0 \times {10^{ - 5}}$ and from equation $ - (1)$we get
                                          ${\alpha _1} = \sqrt {\dfrac{{36.0 \times {{10}^{ - 5}}}}{C}} $ $ - (2)$
And given for acetic acid ${K_a} = 1.8 \times {10^{ - 5}}$ and from equation $ - (1)$ we get
                                            ${\alpha _2} = \sqrt {\dfrac{{1.8 \times {{10}^{ - 5}}}}{C}} $ $ - (3)$
Now, Relative strength is the ratio of degree of dissociation of both acids. So, it can be given as :
                                              $R = \dfrac{{{\alpha _1}}}{{{\alpha _2}}}$ $ - (4)$
Where $R = $ Relative strength
${\alpha _1} = $ Degree of dissociation of m-nitrobenzene
${\alpha _2} = $ Degree of dissociation of acetic acid
Now, by putting equation $ - (2)$ and $ - (3)$in equation $ - (4)$ we get,
$
  R = \sqrt {\dfrac{{36.0 \times {{10}^{ - 5}} \times C}}{{1.8 \times {{10}^{ - 5}} \times C}}} \\
  R = \sqrt {20} \\
  R = 4.47 \\
 $
Hence, we can say that the required relative strength of given acids is $4.47$.

Note:
In such questions, where we need to find relative strength we may get confused as concentration($C$) of acidic solution is not given to find the degree of dissociation constant ($\alpha = \sqrt {\dfrac{{{K_a}}}{C}} $) but we need to remember that we will take concentration of acidic solution($C$) same for two different acids when we are calculating their relative strength.