Question

At $\text{2}{{\text{5}}^{\text{o}}}\text{C}$, the dissociation constant of a base, BOH is $1.0\times {{10}^{-12}}$. The concentration of hydroxyl ions in 0.01 M aqueous solution of the base would be:
(A) $\text{2}\text{.0 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-6}}}\text{mol}{{\text{L}}^{\text{-1}}}$
(B) $\text{1}\text{.0 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-5}}}\text{mol}{{\text{L}}^{\text{-1}}}$
(C) $\text{1}\text{.0 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-6}}}\text{mol}{{\text{L}}^{\text{-1}}}$
(D) $\text{1}\text{.0 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-7}}}\text{mol}{{\text{L}}^{\text{-1}}}$

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. ${{K}_{b}}$ will be equal to the equilibrium constant calculated.
Formula: ${{\text{K}}_{c}}\text{=}\dfrac{\text{Concentration of products}}{\text{Concentration of reactants}}$

Complete step by step answer:
- Let us write down the dissociation of a monoacidic base say BOH,
\[BOH\xrightarrow{reversible}\text{ }{{B}^{+}}+O{{H}^{-}}\]
- From the above dissociation we will now calculate the equilibrium constant for the reaction.
\[\begin{align}
 & \underset{\text{Initial concentration}:\text{ }}{\mathop{{}}}\,\text{ }\underset{\text{c}}{\mathop{\text{BOH}}}\,\text{ }\xrightarrow{\text{reversible}}\text{ }\underset{0}{\mathop{{{\text{B}}^{\text{+}}}}}\,\text{ + }\underset{0}{\mathop{\text{O}{{\text{H}}^{-}}}}\, \\
 & \underset{\text{Final concentration}:\text{ }}{\mathop{{}}}\,\text{ }\underset{\text{c(1-}\alpha \text{)}}{\mathop{\text{BOH}}}\,\text{ }\xrightarrow{\text{reversible}}\text{ }\underset{c\alpha }{\mathop{{{\text{B}}^{\text{+}}}}}\,\text{ + }\underset{c\alpha }{\mathop{\text{O}{{\text{H}}^{-}}}}\, \\
 & \\
\end{align}\]
Where,
$\alpha $ stands for a degree of dissociation.
- We will now substitute the values in the above formula,
\[\begin{align}
 & {{\text{K}}_{b}}\text{=}\frac{\text{Concentration of products}}{\text{Concentration of reactants}} \\
 & {{\text{K}}_{b}}\text{=}\frac{(0.1\alpha )(0.1\alpha )}{0.1(1-\alpha )} \\
\end{align}\]
- The value of ${{K}_{b}}$ is given as $1.0\times {{10}^{-12}}$ and the molarity of solution (c) is 0.01 M. Substituting the values in the equation we get,
\[\begin{align}
 & \left[ \text{O}{{\text{H}}^{\text{-}}} \right]\text{=c }\!\!\alpha\!\!\text{ } \\
 & \left[ O{{H}^{-}} \right]=\sqrt{{{K}_{b}}C}=\sqrt{1\times {{10}^{-12}}\times {{10}^{-2}}} \\
 & =\sqrt{{{K}_{b}}C}=\sqrt{1\times {{10}^{-12}}\times {{10}^{-2}}} \\
 & \left[ \text{O}{{\text{H}}^{\text{-}}} \right]\text{=1}{{\text{0}}^{\text{-7}}}\text{mol}{{\text{L}}^{\text{-1}}} \\
\end{align}\] The correct answer is option “D” .

Additional Information :
- Chemical equilibrium refers to the state of a system in which the concentration of the reactant and the concentration of the products do not change with time and the system does not display any further change in properties.
- When the rate of forward reaction is equal to the rate of the reverse reaction, the state of chemical equilibrium is achieved by the system. When there is no further change in the concentrations of the reactants and the products due to the equal rates of the forward and backward reactions, the system is said to be in a state of dynamic equilibrium.

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