Question

The $ {\text{BOH}} $ is a weak base. Molar concentration of $ {\text{BOH}} $ that provides a $ \left[ {{\text{O}}{{\text{H}}^ - }} \right] $ of $ 1.5 \times {10^{ - 3}}{\text{ M}} $ is:
Given: $ {{\text{K}}_{\text{b}}} $ of $ \left[ {{\text{BOH}}} \right] = 1.5 \times {10^{ - 5{\text{ }}}}{\text{M}} $
(A) $ 0.15{\text{ M}} $
(B) $ 0.1515{\text{ M}} $
(C) $ 0.0015{\text{ M}} $
(D) $ 15 \times {10^{ - 5}}{\text{ M}} $

Answer 1

Hint: To answer this question, you must recall the formula for dissociation constant. Dissociation constant is a modified form of equilibrium constant and it is used to measure the tendency of a compound to break down into its constituent elements or ions or smaller compounds. It is also known as the ionization constant.
Formula used: For the dissociation of weak base $ {\text{BOH}} $ , the equilibrium constant of the process, or the dissociation constant can be written as, $ {{\text{K}}_{\text{b}}} = \dfrac{{\left[ {{{\text{B}}^ + }} \right]\left[ {{\text{O}}{{\text{H}}^ - }} \right]}}{{\left[ {{\text{BOH}}} \right]}} $
Where, $ \left[ {{\text{O}}{{\text{H}}^ - }} \right] $ represents the concentration of hydroxide ion at equilibrium
 $ \left[ {{{\text{B}}^ + }} \right] $ represents the concentration of conjugate acid at equilibrium
And, $ \left[ {{\text{BOH}}} \right] $ represents the concentration of an undissociated base at equilibrium.

Complete step by step solution:
The dissociation of weak base $ {\text{BOH}} $ in aqueous medium can be written as:
 $ {\text{BOH}}\left( {{\text{aq}}{\text{.}}} \right) \rightleftharpoons {{\text{B}}^ + }\left( {{\text{aq}}{\text{.}}} \right) + {\text{O}}{{\text{H}}^ - }\left( {{\text{aq}}{\text{.}}} \right) $
From the reaction, we can see that the amount of the anion and cation produced will be equal, so, we can write, $ \left[ {{{\text{B}}^{\text{ + }}}} \right] = \left[ {{\text{O}}{{\text{H}}^ - }} \right] $
We know the formula for dissociation constant as,
 $ {{\text{K}}_{\text{b}}} = \dfrac{{\left[ {{{\text{B}}^ + }} \right]\left[ {{\text{O}}{{\text{H}}^ - }} \right]}}{{\left[ {{\text{BOH}}} \right]}} = \dfrac{{{{\left[ {{\text{O}}{{\text{H}}^ - }} \right]}^2}}}{{\left[ {{\text{BOH}}} \right]}} $
Substituting the values, we get,
 $ 1.5 \times {10^{ - 5{\text{ }}}}{\text{M}} = \dfrac{{{{\left( {1.5 \times {{10}^{ - 3}}{\text{ M}}} \right)}^2}}}{{\left[ {{\text{BOH}}} \right]}} $
 $ \Rightarrow \left[ {{\text{BOH}}} \right] = 0.15{\text{ M}} $
Thus, the correct answer is A.

Note:
The numerical calculation for equilibrium constant is done by allowing the chemical reaction to reach the equilibrium and then measuring the concentrations of each constituent at that point. Since the concentrations of these constituents are measured at equilibrium point, thus equilibrium constant will always have the same value for a given reaction no matter what is the initial concentration of the reactants.
Using this information about the equilibrium constant, we can derive a standard expression that serves as a standard model for all reactions. This form of the equilibrium constant can be modified as per the requirement of the chemical reaction and is given as
 $ {K_{eq}} = \dfrac{{{{\left( {{a_C}} \right)}^c}}}{{{{\left( {{a_A}} \right)}^a}{{\left( {{a_B}} \right)}^b}}} $ for a reaction: $ aA + bB \rightleftharpoons cC $ .