Question

Assuming that \[{\text{100 cc}}{\text{.}}\] of \[{\text{0}}{\text{.1M}}\] the solution \[{H_2}S{O_4}\] is needed to neutralize \[{\text{200 cc}}\] of solution the normality of \[Ba{\left( {OH} \right)_2}\] a solution is:
A.\[{\text{0}}{\text{.01N}}\]
B.\[{\text{0}}{\text{.5N}}\]
C.\[{\text{0}}{\text{.1N}}\]
D.\[{\text{0}}{\text{.05N}}\]

Answer 1

Hint: To answer this question, you should recall the concept of normality and neutralization reaction. When equal equivalents of an acid and a base are mixed it leads to a neutral solution.
The formula used:
 ${\text{Normality = Molarity}} \times {\text{n - factor}}$

Complete step by step answer:
In the question, it is mentioned that Molarity of \[{H_2}S{O_4}\] \[{\text{ = 0}}{\text{.1 M}}\].
n-factor of \[{H_2}S{O_4}\] = 2 (As seen from the formula that it has 2 dissociable\[{\text{H}}\]).
So, Normality of \[{H_2}S{O_4}\] \[ = {\text{ }}0.1 \times 2{\text{ }}N{\text{ }} = {\text{ }}0.2{\text{ }}N\]
At equivalence point Gram equivalents of \[{H_2}S{O_4}\] = Gram equivalents of \[Ba{\left( {OH} \right)_2}\]
So we can conclude that:
\[{\text{Normalit}}{{\text{y}}_{{H_2}S{O_4}}} \times {\text{Volum}}{{\text{e}}_{{H_2}S{O_4}}} = {\text{Normalit}}{{\text{y}}_{Ba{{\left( {OH} \right)}_2}}} \times {\text{Volum}}{{\text{e}}_{Ba{{\left( {OH} \right)}_2}}}\].
Given that:
\[{\text{Normalit}}{{\text{y}}_{{H_2}S{O_4}}} = 0.2{\text{N}}\], \[{\text{Volum}}{{\text{e}}_{{H_2}S{O_4}}} = 100{\text{cc}}\]and \[{\text{Volum}}{{\text{e}}_{Ba{{\left( {OH} \right)}_2}}} = 200{\text{cc}}\].
Substituting these values in the above equation we get:
$ \Rightarrow 0.2 \times 100 = {\text{Normalit}}{{\text{y}}_{Ba{{\left( {OH} \right)}_2}}} \times 200$.
We will get \[{\text{Normalit}}{{\text{y}}_{Ba{{\left( {OH} \right)}_2}}} = 0.1{\text{N}}\].

Hence, the correct answer to this question is option C.

Note:
You should know about the other concentration terms commonly used:
Concentration in Parts Per Million (ppm) The parts of a component per million parts (\[{10^6}\]) of the solution.
\[{\text{ppm(A)}} = \dfrac{{{\text{Mass of A}}}}{{{\text{Total mass of the solution}}}} \times {10^6}\]
Molality (m): Molality establishes a relationship between moles of solute and the mass of solvent. It is given by moles of solute dissolved per kg of the solvent. The molality formula is as given- \[{\text{Molality(m) = }}\dfrac{{{\text{Mole of solute}}}}{{{\text{Mass of solvent in kg}}}}\]