Solveeit Logo
Login

Question

Question: The pH of 0.1 molar solution of the acid HQ is 3. The value of the ionization constant \({{K}_{a}}\)...

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

Explanation

Solution

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. Ka{{K}_{a}} will be equal to the equilibrium constant calculated.
Formula: Kc=ConcentrationnofproductsConcentrationnofreactants{{\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,
HQ → H+ + Q-\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.
 initial concentration final concentration  HQ c c(1−α)  →reversible    H+ 0 cα  OH− 0 cα  \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,
α\alpha stands for a degree of dissociation.
We will now substitute the values in the above formula,
Ka=Concentration of productsConcentration of reactants{{\text{K}}_{a}}\text{=}\dfrac{\text{Concentration of products}}{\text{Concentration of reactants}}
Ka=(0.1α)(0.1α)0.1(1−α){{\text{K}}_{a}}\text{=}\dfrac{(0.1\alpha )(0.1\alpha )}{0.1(1-\alpha )}
The value of pH given to us is 3.
pH=−log⁡([H+]) 3=−log⁡([H+]) [H+]=antilog(-3)  !![!! H+]=10−3 \begin{aligned} & pH=-\log ([{{H}^{+}}]) \\\ & 3=-\log ([{{H}^{+}}]) \\\ & [{{H}^{+}}]=\text{antilog(-3)} \\\ & \text{ }\\!\\![\\!\\!\text{ }{{\text{H}}^{+}}]={{10}^{-3}} \\\ \end{aligned}
0.1α= 10−3 α=10−2 \begin{aligned} & 0.1\alpha =\text{ 1}{{\text{0}}^{-3}} \\\ & \alpha =\text{1}{{\text{0}}^{-2}} \\\ \end{aligned}
Substituting the value of α\alpha in the equation,
Ka=(0.1α)(0.1α)0.1(1−α){{\text{K}}_{a}}\text{=}\dfrac{(0.1\alpha )(0.1\alpha )}{0.1(1-\alpha )}
We get,
Ka=0.1α2 Ka=10−5 \begin{aligned} & {{\text{K}}_{a}}=0.1{{\alpha }^{2}} \\\ & {{\text{K}}_{a}}={{10}^{-5}} \\\ \end{aligned}
The value of the ionization constant Ka{{K}_{a}} of this acid is 1 x 10−5{{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 = pKw{{K}_{w}} = 14.
Where, Kw{{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{{10}^{-14}} considering the temperature to be 25oC{{25}^{o}}C . When the temperature is increased to 100oC{{100}^{o}}C
the ionic product of water becomes 10−12{{10}^{-12}}.

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