Question

Acetic acid and propionic acid have $ {{K}_{a}} $ values $ 1.75\times {{10}^{-5}} $ and $ 1.3\times {{10}^{-5}} $ respectively at a certain temperature. An equimolar solution of a mixture of the two acids is partially neutralised by $ NaOH\; $ . How is the ratio of the contents of acetate and propionate ions related to the $ {{K}_{a}} $ values and the molarity?
(A) $ \left( \dfrac{\alpha }{1-\alpha } \right)=\dfrac{1.75}{1.3}\times \left( \dfrac{\beta }{1-\beta } \right) $ , where $ \alpha $ and $ \beta $ are ionized fractions of the acids
(B) The ratio is unrelated to the $ {{K}_{a}} $ values
(C) The ratio is unrelated to the molarity
(D) The ratio is unrelated to the $ pH\; $ of the solution

Answer 1

Hint: The value of dissociation constant or ionisation constant $ {{K}_{a}} $ for weak acid can be calculated by taking the ratio of the product of the concentration of the dissociated ions at equilibrium to the concentration of acid at equilibrium. Hence, ionisation constant depends on the molarity of the solution, degree of dissociation, and the $ pH\; $ of the solution .

Complete answer:
The chemical formula for acetic acid is $ C{{H}_{3}}COOH $ and the chemical formula for propionic acid is $ {{C}_{2}}{{H}_{5}}COOH $
Now, the dissociation of both acids in the respective ions can be expressed as
 $ C{{H}_{3}}COOH\to C{{H}_{3}}CO{{O}^{-}}+{{H}^{+}} $
 $ {{C}_{2}}{{H}_{5}}COOH\to {{C}_{2}}{{H}_{5}}CO{{O}^{-}}+{{H}^{+}} $
We can see that the hydrogen ion is obtained in both reactions. Hence, we will consider the total concentration for the ionisation constant.
Let, the degree of ionisation for acetic acid be $ \alpha $ and the degree of ionisation for propionic acid be $ \beta $ and as per the given conditions, equal molarity of both acids as $ C $
Now, let us make an ICE (Initial Change Equilibrium) table for acetic acid

	$ C{{H}_{3}}COOH\to $	$ C{{H}_{3}}CO{{O}^{-}} $	$ {{H}^{+}} $
Initial concentration in Molarity	$ C $	$ 0 $	$ 0 $
Change in concentration or degree of ionisation	$ 1-\alpha $	$ \alpha $	$ \alpha+\beta $
Equilibrium condition	$ C(1-\alpha) $	$ C\alpha $	$ C(\alpha+\beta) $

Now, from the formula for Ionisation constant, Ionisation constant for acetic acid is given as
 $ {{K}_{acetic}}=\dfrac{\left[ C{{H}_{3}}CO{{O}^{-}} \right]\left[ {{H}^{+}} \right]}{\left[ C{{H}_{3}}COOH \right]} $
 $ {{K}_{acetic}}=\dfrac{C\alpha \times C(\alpha +\beta )}{C(1-\alpha )} $ …… $ (1) $
Similarly, the ICE table for the propionic acid is shown as,

	$ {{C}_{2}}{{H}_{5}}COOH\to $	$ {{C}_{2}}{{H}_{5}}CO{{O}^{-}} $	$ {{H}^{+}} $
Initial concentration in Molarity	$ C $	$ 0 $	$ 0 $
Change in concentration or degree of ionisation	$ 1-\beta $	$ \beta $	$ \alpha+\beta $
Equilibrium condition	$ C(1-\beta) $	$ C\beta $	$ C(\alpha+\beta) $

Now, from the formula for Ionisation constant, Ionisation constant for propionic acid is given as
 $ {{K}_{propionic}}=\dfrac{\left[ {{C}_{2}}{{H}_{5}}CO{{O}^{-}} \right]\left[ {{H}^{+}} \right]}{\left[ {{C}_{2}}{{H}_{5}}COOH \right]} $
 $ {{K}_{propionic}}=\dfrac{C\beta \times C(\alpha +\beta )}{C(1-\beta )} $ …… $ (2) $
Taking the ratio of Equation of $ (1) $ and $ (2) $
 $ \dfrac{{{K}_{acetic}}}{{{K}_{propionic}}}=\dfrac{C\alpha \times C(\alpha +\beta )}{C(1-\alpha )}\times \dfrac{C(1-\beta )}{C\beta \times C(\alpha +\beta )} $
Canceling the common terms
 $ \dfrac{{{K}_{acetic}}}{{{K}_{propionic}}}=\dfrac{\alpha }{(1-\alpha )}\times \dfrac{(1-\beta )}{\beta } $
Substituting the given values of ionisation constant
 $ \dfrac{1.75\times {{10}^{-5}}}{1.3\times {{10}^{-5}}}=\dfrac{\alpha }{(1-\alpha )}\times \dfrac{(1-\beta )}{\beta } $
Rearranging the equation,
 $ \therefore \left( \dfrac{\alpha }{1-\alpha } \right)=\dfrac{1.75}{1.3}\times \left( \dfrac{\beta }{1-\beta } \right) $
Hence, the correct answer is Option $ (A) $ .

Note:
Here, the ratio of the ionisation constants $ {{K}_{a}} $ appears to be only proportional to the degree of dissociation of both acids. But, it also depends on the concentration in molarity of the acid. The only reason concentration is not seen in the ratio, is because we are given that both solutions are equimolar, meaning they have the same concentration. Hence, only because of the given conditions, concentration is not present. Else Ionisation constant $ {{K}_{a}} $ , molarity of acid and degree of dissociation are interrelated.