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Question: \({{K}_{b}}\) for a monoacidic base whose 0.1M solution has a pH of 10.50 is:...

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

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. Kb{{K}_{b}} will be equal to the equilibrium constant calculated.
Formula: Kc=ConcentrationofproductsConcentrationofreactants{{\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,
BOHleftrightarrowsreversibleB++OH−BOH{{\overset{reversible}{leftrightarrows}}_{{}}}{{B}^{+}}+O{{H}^{-}}
From the above dissociation we will now calculate the equilibrium constant for the reaction.
 BOH →reversible  B+ + OH− Initial concentration: 0.1 0 0 Final concentration: 0.1(1-α) 0.1α 0.1α \begin{aligned} & \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{aligned}
Where,
α\alpha stands for a degree of dissociation.
We will now substitute the values in the above formula,
Kb=ConcentrationofproductsConcentrationofreactants{{\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}}
Kb=(0.1α)(0.1α)0.1(1−α){{\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.
pH=−log⁡([H+]) 10.5=−log⁡([H+]) [H+]=antilog(-10.5)  !![!! H+]=3.16x10−11 \begin{aligned} & 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{aligned}
The ionic product of water is given by,
[H+][OH−]=10−14[{{H}^{+}}][O{{H}^{-}}]={{10}^{-14}}
Substituting the value of  !![!! H+]\text{ }\\!\\![\\!\\!\text{ }{{\text{H}}^{+}}] in the above equation to calculate the value of [OH−][O{{H}^{-}}],
[3.16x10−11][OH−]=10−14 [OH−]=3.16x10−4 \begin{aligned} & [3.16\text{x1}{{\text{0}}^{-11}}][O{{H}^{-}}]={{10}^{-14}} \\\ & [O{{H}^{-}}]=3.16\text{x1}{{\text{0}}^{-4}} \\\ \end{aligned}
We will now equate the concentration of [OH−][O{{H}^{-}}] to the concentration of [OH−][O{{H}^{-}}] in the dissociation reaction.
0.1α=3.16x10−4 α=3.16x10−3 \begin{aligned} & 0.1\alpha =3.16\text{x1}{{\text{0}}^{-4}} \\\ & \alpha =3.16\text{x1}{{\text{0}}^{-3}} \\\ \end{aligned}
Substituting the value of α\alpha in the equation,
Kb=(0.1α)(0.1α)0.1(1−α){{\text{K}}_{b}}\text{=}\dfrac{(0.1\alpha )(0.1\alpha )}{0.1(1-\alpha )}
We get,
Kb=0.1α2 Kb=10−6 \begin{aligned} & {{\text{K}}_{b}}=0.1{{\alpha }^{2}} \\\ & {{\text{K}}_{b}}={{10}^{-6}} \\\ \end{aligned}

Therefore, the correct answer is option (B).

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.
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}}.