Question

Name the law or principle confirmed by the following observations:
When water is added to 0.01M aqueous solution of acetic acid the number of hydrogen ions increases.

Answer 1

Hint: What kind of electrolyte is acetic acid? Does it dissociate completely? When water is added to any solution, its volume increases and so does the dilution.

Complete answer:
-Acetic acid is a weak acid. It doesn’t dissociate completely in water. It dissociates partially in water. It can be represented as,
\[C{{H}_{3}}COO{{H}_{(aq)}}\rightleftharpoons C{{H}_{3}}COO_{(aq)}^{-}+H_{(aq)}^{+}\]
-Ostwald’s dilution law explains about the dissociation of weak acids.
A weak acid exists in solution partially as ions and partially as an undissociated species.
\[C{{H}_{3}}COO{{H}_{(aq)}}\rightleftharpoons C{{H}_{3}}COO_{(aq)}^{-}+H_{(aq)}^{+}\]
The acid-dissociation constant, ${{K}_{a}}=\dfrac{\left[ C{{H}_{3}}CO{{O}^{-}} \right]\left[ {{H}^{+}} \right]}{\left[ C{{H}_{3}}COOH \right]}$
Let’s say, 1 mole of acid is present initially in volume V $\text{d}{{\text{m}}^{\text{3}}}$ of the solution and the degree of dissociation is $\alpha $. The dfraction of undissociated acid is $(1-\alpha )$.

	\[C{{H}_{3}}COO{{H}_{(aq)}}\rightleftharpoons C{{H}_{3}}COO_{(aq)}^{-}+H_{(aq)}^{+}\]
Amount present at equilibrium	$(1-\alpha )$	$\alpha $	$\alpha $
Concentration at equilibrium $(mol/d{{m}^{3}})$	\[\dfrac{(1-\alpha )}{V}\]	$\dfrac{\alpha }{V}$	$\dfrac{\alpha }{V}$

Substituting these values in the equation ${{K}_{a}}=\dfrac{\left[ C{{H}_{3}}CO{{O}^{-}} \right]\left[ {{H}^{+}} \right]}{\left[ C{{H}_{3}}COOH \right]}$ we get,
${{K}_{a}}=\dfrac{\left( \dfrac{\alpha }{V} \right)\left( \dfrac{\alpha }{V} \right)}{\left( \dfrac{1-\alpha }{V} \right)}=\dfrac{{{\alpha }^{2}}}{\left( 1-\alpha \right)V}$
If C is the initial concentration of acid, then $C={}^{1}/{}_{V}$. Therefore, we get, ${{K}_{a}}=\dfrac{{{\alpha }^{2}}C}{1-\alpha }$
For a weak acid, $\alpha $ is very small. So, $(1-\alpha )\approx 1$.
So, the equation becomes, ${{K}_{a}}={{\alpha }^{2}}C$ or ${{K}_{a}}={}^{{{\alpha }^{2}}}/{}_{V}$
Hence, $\alpha =\sqrt{{{K}_{a}}V}$ or $\alpha =\sqrt{{}^{{{K}_{a}}}/{}_{C}}$
This equation implies that the degree of dissociation of a weak acid is inversely proportional to the square root of its concentration or directly proportional to the square root of volume of the solution. That means, as the volume increases, more is the dissociation of weak acid.

When water is added to 0.01M aqueous solution of acetic acid the number of hydrogen ions increases. This statement is proved using Ostwald’s dilution law.

Note:
This question comes under ionic equilibria chapter. Please don’t get confused with the molar conductivity concept which is a part of electrochemistry which says conductivity decreases with increase in dilution. Here, in ionic equilibria, for weak electrolytes, degree of dissociation increases with increase in dilution.