Question

${{K}_{b}}$ for a monoacidic base whose 0.1M solution has a pH of 10.50 is:

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: ${{\text{K}}_{\text{c}}}\text{=}\dfrac{\text{Concentratio}{{\text{n}}_{{}}}\text{o}{{\text{f}}_{{}}}\text{products}}{\text{Concentratio}{{\text{n}}_{{}}}\text{o}{{\text{f}}_{{}}}\text{reactants}}$

Complete answer:
Let us write down the dissociation of a monoacidic base say BOH,
$BOH{{\overset{reversible}{leftrightarrows}}_{{}}}{{B}^{+}}+O{{H}^{-}}$
From the above dissociation we will now calculate the equilibrium constant for the reaction.
     \[\begin{align}
  & \text{ BOH }{{\overset{\text{reversible}}{\mathop{\xrightarrow{{}}}}\,}_{{}}}\text{ }{{\text{B}}^{\text{+}}}\text{ + O}{{\text{H}}^{-}} \\
 & \text{Initial concentration}:\text{ 0}\text{.1 0 0} \\
 & \text{Final concentration: 0}\text{.1(1-}\alpha \text{) 0}\text{.1}\alpha \text{ 0}\text{.1}\alpha \\
\end{align}\]
Where,
$\alpha $stands for a degree of dissociation.
We will now substitute the values in the above formula,
${{\text{K}}_{b}}\text{=}\dfrac{\text{Concentratio}{{\text{n}}_{{}}}\text{o}{{\text{f}}_{{}}}\text{products}}{\text{Concentratio}{{\text{n}}_{{}}}\text{o}{{\text{f}}_{{}}}\text{reactants}}$
${{\text{K}}_{b}}\text{=}\dfrac{(0.1\alpha )(0.1\alpha )}{0.1(1-\alpha )}$
The value of pH given to us is 10.5.
$\begin{align}
  & pH=-\log ([{{H}^{+}}]) \\
 & 10.5=-\log ([{{H}^{+}}]) \\
 & [{{H}^{+}}]=\text{antilog(-10}\text{.5)} \\
 & \text{ }\!\![\!\!\text{ }{{\text{H}}^{+}}]=3.16\text{x1}{{\text{0}}^{-11}} \\
\end{align}$
The ionic product of water is given by,
$[{{H}^{+}}][O{{H}^{-}}]={{10}^{-14}}$
Substituting the value of $\text{ }\!\![\!\!\text{ }{{\text{H}}^{+}}]$ in the above equation to calculate the value of $[O{{H}^{-}}]$,
$\begin{align}
  & [3.16\text{x1}{{\text{0}}^{-11}}][O{{H}^{-}}]={{10}^{-14}} \\
 & [O{{H}^{-}}]=3.16\text{x1}{{\text{0}}^{-4}} \\
\end{align}$
We will now equate the concentration of $[O{{H}^{-}}]$ to the concentration of $[O{{H}^{-}}]$ in the dissociation reaction.
$\begin{align}
  & 0.1\alpha =3.16\text{x1}{{\text{0}}^{-4}} \\
 & \alpha =3.16\text{x1}{{\text{0}}^{-3}} \\
\end{align}$
Substituting the value of $\alpha $in the equation,
${{\text{K}}_{b}}\text{=}\dfrac{(0.1\alpha )(0.1\alpha )}{0.1(1-\alpha )}$
We get,
$\begin{align}
  & {{\text{K}}_{b}}=0.1{{\alpha }^{2}} \\
 & {{\text{K}}_{b}}={{10}^{-6}} \\
\end{align}$

Therefore, the correct answer is option (B).

Note:
In the above calculation we have ignored the value$(1-\alpha )$ in the denominator. This is because the value of $\alpha \ll 1$.
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 ${{10}^{-12}}$.