Question

The pH of 0.1 molar solution of the acid HQ is 3. The value of the ionization constant ${{K}_{a}}$ of this acid is:
(A) 3 x ${{10}^{-1}}$
(B) 1 x ${{10}^{-3}}$
(C) 1 x ${{10}^{-5}}$
(D) 1 x ${{10}^{-7}}$

Answer 1

Hint: Write down the dissociation of the base in water. Find the equilibrium constant for the dissociation process with the degree of dissociation. The formula for equilibrium constant is given below. ${{K}_{a}}$ will be equal to the equilibrium constant calculated.
Formula: ${{\text{K}}_{\text{c}}}\text{=}\dfrac{\text{Concentration}{{\text{n}}_{{}}}\text{o}{{\text{f}}_{{}}}\text{products}}{\text{Concentration}{{\text{n}}_{{}}}\text{o}{{\text{f}}_{{}}}\text{reactants}}$

Complete step-by-step answer:

Let us write down the dissociation of a monoacidic base say HQ,
$\text{HQ }\to \text{ }{{\text{H}}^{\text{+}}}\text{ + }{{\text{Q}}^{\text{-}}}$
From the above dissociation we will now calculate the equilibrium constant for the reaction.
$\begin{matrix}
   {} \\
   initial\ concentration \\
   final\ concentration \\
\end{matrix}\text{ }\begin{matrix}
   HQ \\
   c \\
   c(1-\alpha ) \\
\end{matrix}\text{ }\begin{matrix}
   \xrightarrow{reversible} \\
   {} \\
   {} \\
\end{matrix}\text{ }\begin{matrix}
   {{H}^{+}} \\
   0 \\
   c\alpha \\
\end{matrix}\text{ }\begin{matrix}
   O{{H}^{-}} \\
   0 \\
c\alpha \\
\end{matrix}\text{ }$
Where,
$\alpha $stands for a degree of dissociation.
We will now substitute the values in the above formula,
${{\text{K}}_{a}}\text{=}\dfrac{\text{Concentration of products}}{\text{Concentration of reactants}}$
${{\text{K}}_{a}}\text{=}\dfrac{(0.1\alpha )(0.1\alpha )}{0.1(1-\alpha )}$
The value of pH given to us is 3.
$\begin{align}
  & pH=-\log ([{{H}^{+}}]) \\
 & 3=-\log ([{{H}^{+}}]) \\
 & [{{H}^{+}}]=\text{antilog(-3)} \\
 & \text{ }\!\![\!\!\text{ }{{\text{H}}^{+}}]={{10}^{-3}} \\
\end{align}$
$\begin{align}
& 0.1\alpha =\text{ 1}{{\text{0}}^{-3}} \\
& \alpha =\text{1}{{\text{0}}^{-2}} \\
\end{align}$
Substituting the value of $\alpha $in the equation,
${{\text{K}}_{a}}\text{=}\dfrac{(0.1\alpha )(0.1\alpha )}{0.1(1-\alpha )}$
We get,
$\begin{align}
& {{\text{K}}_{a}}=0.1{{\alpha }^{2}} \\
& {{\text{K}}_{a}}={{10}^{-5}} \\
\end{align}$
The value of the ionization constant ${{K}_{a}}$ of this acid is 1 x ${{10}^{-5}}$.

Therefore, the correct answer is option (C).

Additional information: The ionic product of water is mainly used for calculating pOH value when pH is given to us or vice versa.
The relation between the three quantities is,
pH + pOH = p${{K}_{w}}$ = 14.
Where, ${{K}_{w}}$ is defined as the ionic product of water.
The ionic product of water depends primarily on temperature .The ionic product of water is ${{10}^{-14}}$ considering the temperature to be ${{25}^{o}}C$ . When the temperature is increased to ${{100}^{o}}C$
the ionic product of water becomes ${{10}^{-12}}$.

Note: In the above calculation we have ignored the value$(1-\alpha )$ in the denominator. This is because the value of $\alpha \ll 1$.