Question

For the standardization of $ {\text{Ba}}{({\text{OH}})_2} $ solution, $ 0.204\;{\text{g}} $ of potassium acid phthalate was weighed which was then titrated with $ {\text{Ba}}{({\text{OH}})_2} $ solution. The titration indicated equivalence at $ 25.0{\text{ml}} $ of $ {\text{Ba}}{({\text{OH}})_2} $ solution. The reaction involved is:
 $ {\text{KH}}{{\text{C}}_8}{{\text{H}}_4}{{\text{O}}_4} + {\text{Ba}}{({\text{OH}})_2} \to {{\text{H}}_2}{\text{O}} + {{\text{K}}^ + } + {\text{B}}{{\text{a}}^{2 + }} + {{\text{C}}_8}{{\text{H}}_4}{\text{O}}_4^{2 - } $
The molarity of the base solution is $ ({\text{K}} = 39) $
(A) $ 0.04{\text{M}} $
(B) $ 0.03{\text{M}} $
(C) $ 0.02{\text{M}} $
(D) $ 0.01{\text{M}} $

Answer 1

Hint: A neutralization reaction can be defined as a chemical reaction in which an acid and a base react together quantitatively to form a product of salt and water. The above reaction in the question is also a neutralization reaction. We shall calculate the normality of the solution using the formula given and then convert it to molarity.

Formula used:
 $ {\text{normality}} = {\text{molarity }} \times {\text{n - factor}} $
And
 $ {{\text{N}}_1}\;{{\text{V}}_1} = {{\text{N}}_2}\;{{\text{V}}_2} $
Where
 $ {{\text{N}}_1} $ is the normality of potassium acid phthalate $ {\text{KH}}{{\text{C}}_8}{{\text{H}}_4}{{\text{O}}_4} $
 $ {{\text{V}}_1} $ is the volume of $ {\text{KH}}{{\text{C}}_8}{{\text{H}}_4}{{\text{O}}_4} $
 $ {{\text{N}}_2}\; $ is the normality of $ {\text{Ba}}{({\text{OH}})_2} $
 $ \;{{\text{V}}_2} $ is the volume of $ {\text{Ba}}{({\text{OH}})_2} $ .

Complete step by step solution:
The weight of potassium acid phthalate $ {\text{KH}}{{\text{C}}_8}{{\text{H}}_4}{{\text{O}}_4} $ used $ = 0.204\;{\text{g}} $
The above chemical reaction is a neutralization reaction. We know that for complete neutralization,
no. of equivalents of $ \left( {{\text{KH}}{{\text{C}}_8}{{\text{H}}_4}{{\text{O}}_4}} \right) = $ no. of equivalents of $ {\text{Ba}}{({\text{OH}})_2} $
Now,
 $ {{\text{N}}_1}\;{{\text{V}}_1} = {{\text{N}}_2}\;{{\text{V}}_2} $
The molecular weight of potassium acid phthalate $ {\text{KH}}{{\text{C}}_8}{{\text{H}}_4}{{\text{O}}_4} = 208{\text{g/mol}} $
Equivalent wt. of $ \left( {{\text{KH}}{{\text{C}}_8}\;{{\text{N}}_4}{{\text{O}}_4}} \right) = \dfrac{{{\text{molecular weight}}}}{{{\text{n}} - {\text{factor}}}} = \dfrac{{204}}{1} = 204 $
Since, the number of replaceable $ {\text{H}} $ is $ 1 $ , then the n-factor will also be equal to $ 1 $ .
 $ {{\text{N}}_2} = \dfrac{{0.204}}{{204 \times 0.025}} $
So, we get
 $ {{\text{N}}_2} = 0.04 N $ $ {\text{Ba}}{({\text{OH}})_2} $
We also know that $ {\text{Ba}}{({\text{OH}})_2} $ is a di-acidic base with n-factor $ = 2 $
We know that
 $ {\text{normality}} = {\text{molarity}} \times {\text{n - factor}} $
Now, we will put the values of normality and n-factor in the above formula to get the molarity of $ {\text{Ba}}{({\text{OH}})_2} $
That is
 $ 0.04 = {\text{molarity}} \times 2 $
Molarity of $ {\text{Ba}}{({\text{OH}})_2} $ $ = \dfrac{{0.04}}{2} = 0.02 M $
Hence, the correct option is (C.)

Note:
Normality can be defined as the gram equivalent of a dissolved solution in a single litre. The unit of normality is N. During titration calculations, it is most preferred. Normal solutions have equivalent normality to unity.
The number of moles of solvent dissolved in one litre of solution can be defined as molarity. The unit of molarity is M. It is the preferred concentration unit.