Question

Some amount of \[{\text{N}}{{\text{H}}_{\text{4}}}{\text{Cl}}\] was boiled with \[{\text{50 mL}}\] of \[{\text{0}}{\text{.75 N NaOH}}\] solution till the reaction was complete. After the completion of the reaction, \[{\text{10 mL}}\] of \[{\text{0}}{\text{.75 N }}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}\] were required for the neutralisation of the remaining \[{\text{NaOH}}\] . Calculate the amount of \[{\text{N}}{{\text{H}}_{\text{4}}}{\text{Cl}}\] taken.

Answer 1

Hint:Normality is the number of gram equivalents of solute present in one liter of solution. The number of gram equivalents is the ratio of the mass to the gram equivalent weight of solute. The gram equivalent mass of an acid is equal to the ratio of the molecular weight to basicity of an acid.

Complete step-by-step solution:
Given that
\[{\text{10 mL}}\] of \[{\text{0}}{\text{.75 N }}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}\] will react with \[{\text{10 mL}}\] of \[{\text{0}}{\text{.75 N NaOH}}\].
Thus, out of \[{\text{50 mL}}\] of \[{\text{0}}{\text{.75 N NaOH}}\], the volume that reacts with \[{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}\] is
\[{\text{10 mL}}\] of \[{\text{0}}{\text{.75 N NaOH}}\]and the volume that reacts with \[{\text{N}}{{\text{H}}_{\text{4}}}{\text{Cl}}\] is \[\left( {{\text{50 mL}} - 1{\text{0 mL}}} \right) = {\text{40 mL}}\] of \[{\text{0}}{\text{.75 N NaOH}}\] .
\[{\text{40 mL}}\] of \[{\text{0}}{\text{.75 N NaOH}}\] will react with \[{\text{40 mL}}\] of \[{\text{0}}{\text{.75 N N}}{{\text{H}}_{\text{4}}}{\text{Cl}}\]
Using the following formula, to calculate the mass of \[{\text{N}}{{\text{H}}_{\text{4}}}{\text{Cl}}\] we will get
\[{\text{Mass of N}}{{\text{H}}_{\text{4}}}{\text{Cl = }}\dfrac{{{\text{Normality }} \times {\text{ equivalent mass }} \times {\text{ Volume}}}}{{1000}}\]
Let us Substitute \[{\text{0}}{\text{.75 N}}\] as normality of \[{\text{N}}{{\text{H}}_{\text{4}}}{\text{Cl}}\] , 53.5 as equivalent mass of \[{\text{N}}{{\text{H}}_{\text{4}}}{\text{Cl}}\] and \[{\text{40 mL}}\] as volume of \[{\text{N}}{{\text{H}}_{\text{4}}}{\text{Cl}}\] in the above equation and calculate the mass of \[{\text{N}}{{\text{H}}_{\text{4}}}{\text{Cl}}\]
\[
  {\text{Mass of N}}{{\text{H}}_{\text{4}}}{\text{Cl = }}\dfrac{{{\text{0}}{\text{.75 N }} \times {\text{ 53}}{\text{.5 g/eq }} \times {\text{ 40 mL}}}}{{1000}} \\
   = \dfrac{{{\text{270}}{\text{.3 g }}}}{{1000}} \\
   = 0.2703{\text{ g}} \\
 \]
Hence, the amount of \[{\text{N}}{{\text{H}}_{\text{4}}}{\text{Cl}}\] taken is \[0.2703{\text{ g}}\]


Note:Normality is similar to molarity. In molarity, the number of moles of solute are divided with the volume of the solution. In molarity, the number of gram equivalents are divided by the volume of the solution.