Question

$5{\text{mL}}$ of $8{\text{N}}$ nitric acid, $4.8{\text{mL}}$ of $5{\text{N}}$ hydrochloric acid and a certain volume of $17{\text{M}}$ sulfuric acid are mixed together and made up to $2{\text{L}}$. $30{\text{mL}}$ of this acid mixture exactly neutralize $42.9{\text{mL}}$ of sodium carbonate solution containing $1{\text{g}}$ of ${\text{N}}{{\text{a}}_2}{\text{C}}{{\text{O}}_3}.10{{\text{H}}_2}{\text{O}}$ in $100{\text{mL}}$ of water. Calculate the amount (in ${\text{g}}$ ) of the sulfate ions in solution.

Answer 1

In all the techniques of quantitative analysis, the use of solutions requires some basis for the expression of solution concentration. Normality is the strength of solution measured in terms of gram equivalents. Normality depends on the equivalent mass and volume. Using normality, the volume of the acid can be calculated. With volume and the equivalent weight of the acid, the amount of the sulfate ions is calculated.

Complete step by step answer:
Given that,
Volume of nitric acid, ${{\text{V}}_{{\text{HN}}{{\text{O}}_3}}} = 5{\text{mL}}$
Normality of nitric acid, ${{\text{N}}_{{\text{HN}}{{\text{O}}_3}}} = 8{\text{N}}$
Volume of ${\text{HCl}},{{\text{V}}_{{\text{HCl}}}} = 4.8{\text{mL}}$
Normality of ${\text{HCl}},{{\text{N}}_{{\text{HCl}}}} = 5{\text{N}}$
Molarity of ${{\text{H}}_2}{\text{S}}{{\text{O}}_4},{{\text{M}}_{{{\text{H}}_2}{\text{S}}{{\text{O}}_4}}} = 17{\text{M}}$
Volume of ${\text{N}}{{\text{a}}_2}{\text{C}}{{\text{O}}_3}$ , ${{\text{V}}_{{\text{N}}{{\text{a}}_2}{\text{C}}{{\text{O}}_3}}} = 42.9{\text{mL}}$
Volume of acid mixture, ${{\text{V}}_{{\text{acid}}}} = 30{\text{mL}}$

Three acids of different normality and volumes are mixed together and made up the solution. From this, a certain volume was needed to neutralize sodium carbonate. We need to calculate the amount of sulfate ions.
Molecular mass of ${\text{N}}{{\text{a}}_2}{\text{C}}{{\text{O}}_3}.10{{\text{H}}_2}{\text{O}}$ $= 286$
Valency of sodium carbonate depends upon the number of charges it dissociates, i.e. the valency is $2$.
Hence, equivalent mass of ${\text{N}}{{\text{a}}_2}{\text{C}}{{\text{O}}_3}.10{{\text{H}}_2}{\text{O}}$ $= \dfrac{{286}}{2} = 143$
It is given that $100{\text{mL}}$ sodium carbonate solution contains $1{\text{g}}$
So $1000{\text{mL}}$ sodium carbonate solution contains $\dfrac{{1{\text{g}}}}{{100{\text{mL}}}} \times 1000{\text{mL = 10g}}$
Thus, normality can be calculated.
i.e. Normality $= \dfrac{{{\text{Number}}\;{\text{of}}\;{\text{gram}}\;{\text{equivalents}}}}{{{\text{Volume}}\;{\text{in}}\;{\text{litres}}}}$
Number of gram equivalents $= \dfrac{{{\text{Weight}}\;{\text{of}}\;{\text{solute}}}}{{{\text{Equivalent}}\;{\text{weight}}\;{\text{of}}\;{\text{solute}}}}$
Thus normality $= \dfrac{{{\text{Weight}}\;{\text{of}}\;{\text{solute}}}}{{{\text{Equivalent}}\;{\text{weight}}\;{\text{of}}\;{\text{solute}}}}{\text{ X }}\dfrac{{\text{1}}}{{{\text{Volume}}\;{\text{in}}\;{\text{litres}}}}$
Normality, ${\text{N}} = \dfrac{{10}}{{143}} \times \dfrac{1}{1} = \dfrac{{10}}{{143}}{\text{N}}$
All the three acids are mixed together and a volume of $30{\text{mL}}$ of this mixture neutralized $42.9{\text{mL}}$ of sodium carbonate solution.
Thus, the volumes of acid and ${\text{N}}{{\text{a}}_2}{\text{C}}{{\text{O}}_3}$ is given.
i.e. Volume of acid mixture, ${{\text{V}}_{{\text{acid}}}} = 30{\text{mL}}$
Volume of ${\text{N}}{{\text{a}}_2}{\text{C}}{{\text{O}}_3}$ , ${{\text{V}}_{{\text{N}}{{\text{a}}_2}{\text{C}}{{\text{O}}_3}}} = 42.9{\text{mL}}$
Normality of ${\text{N}}{{\text{a}}_2}{\text{C}}{{\text{O}}_3}$, ${{\text{N}}_{{\text{N}}{{\text{a}}_2}{\text{C}}{{\text{O}}_3}}} = \dfrac{{10}}{{143}}{\text{N}}$
Thus, normality of acid can be calculated using the formula give below:
${{\text{N}}_{{\text{N}}{{\text{a}}_2}{\text{C}}{{\text{O}}_3}}}{{\text{V}}_{_{{\text{N}}{{\text{a}}_2}{\text{C}}{{\text{O}}_3}}}} = {{\text{N}}_{{\text{acid}}}}{{\text{V}}_{{\text{acid}}}}$
Normality of acid, ${{\text{N}}_{{\text{acid}}}} = \dfrac{{{{\text{N}}_{{\text{N}}{{\text{a}}_2}{\text{C}}{{\text{O}}_3}}} \times {{\text{V}}_{{\text{N}}{{\text{a}}_2}{\text{C}}{{\text{O}}_3}}}}}{{{{\text{V}}_{{\text{acid}}}}}}$
Substituting the values, we get
${{\text{N}}_{{\text{acid}}}} = \dfrac{{10 \times 42.9}}{{143 \times 30}} = \dfrac{{429}}{{4290}} = 0.1{\text{N}}$
The mixture contains $5{\text{mL}}$ volume of $8{\text{N}}$nitric acid, $4.8{\text{mL}}$volume of $5{\text{N}}$ hydrochloric acid and a ${\text{V}}$ of $17{\text{M}}$ sulfuric acid.
We can say that the sum of the product of normality and volume of each acid is equal to the product of normality and volume of the mixture.
i.e. ${{\text{N}}_{{\text{HN}}{{\text{O}}_3}}}{{\text{V}}_{{\text{HN}}{{\text{O}}_{\text{3}}}}} + {{\text{N}}_{{\text{HCl}}}}{{\text{V}}_{{\text{HCl}}}} + {{\text{N}}_{{{\text{H}}_2}{\text{S}}{{\text{O}}_4}}}{{\text{V}}_{{{\text{H}}_2}{\text{S}}{{\text{O}}_4}}} = {{\text{N}}_{{\text{mix}}}}{{\text{V}}_{{\text{mix}}}}$
Substituting the values, we get
$\Rightarrow 8\,{\text{N}} \times 5\,{\text{mL}} + 5\,{\text{N}} \times 4.8\,{\text{mL}} + 34\,{\text{N}} \times {\text{V}}\,{\text{mL}} = 0.1\,{\text{N}} \times 2000\,{\text{mL}}$
On simplifying,
$\Rightarrow 40 + 2.4 + 34{\text{V}} = 200$
$\Rightarrow 42.4 + 34{\text{V}} = 200 \\\ \Rightarrow 34{\text{V}} = 200 - 42.4 = 157.6 \\\ \Rightarrow {\text{V}} = \dfrac{{157.6}}{{34}} = 4.6\,{\text{mL}} \\\$
We have to find the amount of sulfate ions in the solution.
For that, we need to know the equivalent weight of sulfuric acid.
Equivalent weight of ${{\text{H}}_2}{\text{S}}{{\text{O}}_4}$ $= \dfrac{{{\text{Molecular}}\;{\text{weight}}\;}}{{{\text{valency}}}} = \dfrac{{98}}{2} = 49$
Amount of ${\text{S}}{{\text{O}}_4}^{2 - }$ $= \dfrac{{{\text{Normality X equivalent}}\;{\text{weight X volume}}}}{{{\text{1000}}}} = \dfrac{{34{\text{N}} \times 49 \times 4.6{\text{mL}}}}{{1000}}$
Simplifying,
Amount of ${\text{S}}{{\text{O}}_4}^{2 - }$ $= \dfrac{{7663.6}}{{1000}} = 7.66{\text{g}}$
Thus, the amount of sulfate ions is $7.66{\text{g}}$.

Additional information:
There are different methods to express concentration. Concentration is the amount of solute dissolved in a given amount of solution. Different units of concentration are molarity, molality, normality, parts per million, billion, trillion, mass percent, percent weight by volume, etc. Inter conversion of units of concentration is also possible.

Note:
Titration, also known as titrimetry, is a common laboratory method of quantitative chemical analysis that is used to determine the unknown concentration of an identified analyte.The most common types of qualitative titration are acid-base titrations and redox titrations.