Question

Calculate the mass of ${\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{C}}{{\rm{O}}_{\rm{3}}}$ required to prepare $250\;{\rm{ml}}$ of $0.5\;{\rm{N}}$solution.

Answer 1

Hint:Normality is defined as the number of gram or mole equivalents of solute present in one litre of a solution. It can be calculated by dividing gram equivalents of solute present in one litre of solution. Here, we have to use formula of normality, that is ${\rm{Normality}} = \dfrac{{{\rm{Gram}}\;{\rm{equivalents}}\;{\rm{of}}\;{\rm{solute}}}}{{{\rm{Volume}}\;{\rm{of}}\;{\rm{solution}}\;{\rm{in}}\;{\rm{litre}}}}$

Complete step by step answer:
To calculate normality, we need the gram equivalents of solute, that is, ${\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{C}}{{\rm{O}}_{\rm{3}}}$.
The formula to calculate gram equivalent of solute is,
${\rm{Gram}}\;{\rm{equivalent}} = \dfrac{{{\rm{Weight}}}}{{{\rm{equivalent}}\;{\rm{weight}}}}$ …… (1)
Now, we calculate the equivalent weight of solute.
${\rm{Equivalent}}\;{\rm{weight}} = \dfrac{{{\rm{Molecular}}\;{\rm{mass}}\;{\rm{of}}\,{\rm{solute}}}}{{n - {\rm{factor}}}}$
So, we need molecular mass and n-factor of solute.
First we calculate the molecular mass of ${\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{C}}{{\rm{O}}_{\rm{3}}}$.
$\begin{array}{c}{\rm{Molecular}}\;{\rm{mass}}\;{\rm{of}}\;{\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{C}}{{\rm{O}}_{\rm{3}}} = 2 \times 23 + 12 + 3 \times 16\\ = 46 + 12 + 48\\ = 106\;{\rm{g}}\,{\rm{mo}}{{\rm{l}}^{ - 1}}\end{array}$
Now, we calculate the n-factor of${\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{C}}{{\rm{O}}_{\rm{3}}}$. The n-factor is defined as the change in oxidation number.
The dissociation of ${\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{C}}{{\rm{O}}_{\rm{3}}}$ is written as follows:
 ${\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{C}}{{\rm{O}}_{\rm{3}}} \to 2{\rm{N}}{{\rm{a}}^ + } + {\rm{C}}{{\rm{O}}_{\rm{3}}}^{2 - }$.
No n-factor for${\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{C}}{{\rm{O}}_{\rm{3}}}$is $ = 2 - 0 = 2$.
Now, we use the n-factor and molecular mass of to ${\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{C}}{{\rm{O}}_{\rm{3}}}$calculate equivalent mass of ${\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{C}}{{\rm{O}}_{\rm{3}}}$.
$\begin{array}{c}{\rm{Equivalent}}\;{\rm{mass}}\;{\rm{of}}\;{\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{C}}{{\rm{O}}_{\rm{3}}} = \dfrac{{{\rm{Molecular}}\;{\rm{mass}}\;{\rm{of}}\;{\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{C}}{{\rm{O}}_{\rm{3}}}}}{{{\rm{n}} - {\rm{factor}}\,{\rm{of}}\;{\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{C}}{{\rm{O}}_{\rm{3}}}}}\\ = \dfrac{{106\;{\rm{g}}\;{\rm{mo}}{{\rm{l}}^{ - 1}}}}{2}\\ = 53\;{\rm{equiv}}\,{\rm{mo}}{{\rm{l}}^{ - 1}}\end{array}$

So, equivalent mass of ${\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{C}}{{\rm{O}}_{\rm{3}}}$is $53\;{\rm{equiv}}\,{\rm{mo}}{{\rm{l}}^{ - 1}}$. Now, we have to use equation (1) to calculate the gram equivalent of ${\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{C}}{{\rm{O}}_{\rm{3}}}$ .
$\begin{array}{c}{\rm{Gram}}\;{\rm{equivalent}} = \dfrac{{{\rm{Weight}}}}{{{\rm{equivalent}}\;{\rm{weight}}}}\\ = \dfrac{{{\rm{Weight}}\,{\rm{of}}\;{\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{C}}{{\rm{O}}_{\rm{3}}}}}{{{\rm{equivalent}}\;{\rm{weight}}\;{\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{C}}{{\rm{O}}_{\rm{3}}}\;}}\\ = \dfrac{{{\rm{Weight}}\,{\rm{of}}\;{\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{C}}{{\rm{O}}_{\rm{3}}}}}{{53\;{\rm{g}}\;{\rm{equi}}{{\rm{v}}^{ - 1}}}}\end{array}$
Now, we have to use the formula of normality to calculate the mass of ${\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{C}}{{\rm{O}}_{\rm{3}}}$. The volume of the solution is given as 250 ml and normality is 0.5 N.
$\begin{array}{c}{\rm{Normality}} = \dfrac{{{\rm{Gram}}\;{\rm{equivalents}}\;{\rm{of}}\;{\rm{solute}}}}{{{\rm{Volume}}\;{\rm{of}}\;{\rm{solution}}\;{\rm{in}}\;{\rm{litre}}}}\\0.5\;{\rm{g}}\;{\rm{equiv}}\,{{\rm{L}}^{ - 1}}\left( {\rm{N}} \right) = \dfrac{{{\rm{Weight}}\,{\rm{of}}\;{\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{C}}{{\rm{O}}_{\rm{3}}}}}{{53\;{\rm{g}}\;{\rm{equi}}{{\rm{v}}^{ - 1}}}} \times \dfrac{{1000}}{{250}}{\rm{L}}\\{\rm{Weight}}\,{\rm{of}}\;{\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{C}}{{\rm{O}}_{\rm{3}}} = \dfrac{{0.5\; \times 53 \times 250}}{{1000}}\\ = 6.625\;{\rm{g}}\end{array}$

Hence, the mass of sodium carbonate needed to prepare 250 ml of 0.5 N solution is 6.625 g.

Note:
Molarity is the moles of solute present in one litre of solution and normality is the gram equivalents of solute present in one litre of solution. For some monovalent compounds, like HCl, molarity and normality is the same as n-factor in such compounds.