Question

1.03 g mixture of sodium carbonate and calcium carbonate require 20 ml N${\text{HCl}}$ for complete neutralisation. Calculate the percentage of sodium carbonate and calcium carbonate in the given mixture.

Answer 1

Write the neutralisation reaction for sodium carbonate and calcium carbonate with ${\text{HCl}}$. Assume the mass of one carbonate as ‘x’ and calculate the mass of another carbonate in terms x using a mass of the mixture. Using these masses of each carbonate and their respective equivalent weights set up the equation for a number of g equivalents for each carbonate. Using normality of ${\text{HCl}}$ and its volume calculate the number of g equivalents of${\text{HCl}}$.Using a number of g equivalent of ${\text{HCl}}$ and carbonates calculate the mass of carbonate that is assumed as x. Finally, calculate the percentage of sodium carbonate and calcium carbonate in the given mixture.

Formula Used:
${\text{Number of g equivalent }} = \dfrac{{{\text{weight}}}}{{{\text{equivalent weight}}}}$
${\text{Number of g equivalent}} = {\text{ Normality }} \times {\text{ volume in L}}$
{\text{ Mass % substance = }}\dfrac{{{\text{ Mass of substance }}}}{{{\text{ Mass of mixture }}}} \times 100\%

Complete step-by-step answer:
We have given a mixture of sodium carbonate and calcium carbonate. This mixture is neutralised using ${\text{HCl}}$. So let us write the neutralisation reaction for sodium carbonate and calcium carbonate with${\text{HCl}}$.

${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}{\text{ + 2HCl }} \to {\text{ 2 NaCl + }}{{\text{H}}_{\text{2}}}{\text{O + C}}{{\text{O}}_{\text{2}}}$

${\text{CaC}}{{\text{O}}_{\text{3}}}{\text{ + 2HCl }} \to {\text{ CaC}}{{\text{l}}_{\text{2}}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{O + C}}{{\text{O}}_{\text{2}}}$

Let us assume the mass of ${\text{CaC}}{{\text{O}}_{\text{3}}}$ as ‘x g’

We have given the mass of the mixture of sodium carbonate and calcium carbonate is 1.03 g.

So, the mass of ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ = mass of mixture – a mass of ${\text{CaC}}{{\text{O}}_{\text{3}}}$

${\text{ mass of N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}} = {\text{ (1}}{\text{.03 - x) g }}$

Now, we will calculate the number of g equivalents for each carbonate.

The molecular weight of ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$= ${\text{106g/mol}}$

Equivalent weight of${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$= $\dfrac{{106}}{2} = {\text{ 53g }}$

${\text{Number of g equivalent of N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}} = \dfrac{{{\text{weight}}}}{{{\text{equivalent weight}}}} = \dfrac{{(1.03 - x){\text{ g}}}}{{53{\text{g}}}}$

Molecular weight of ${\text{CaC}}{{\text{O}}_{\text{3}}}$= ${\text{100g/mol}}$

Equivalent weight of${\text{CaC}}{{\text{O}}_{\text{3}}}$= $\dfrac{{100}}{2} = {\text{ 50g }}$

${\text{Number of g equivalent of CaC}}{{\text{O}}_{\text{3}}} = \dfrac{{{\text{weight}}}}{{{\text{equivalent weight}}}} = \dfrac{{{\text{x g}}}}{{50{\text{g}}}}$

Now, using normality of ${\text{HCl}}$ and its volume we will calculate the number of g equivalents of${\text{HCl}}$ as follows:
${\text{Number of g equivalent of HCl}} = {\text{ Normality }} \times {\text{ volume in L = }}\dfrac{{{\text{1N }} \times {\text{20 ml}}}}{{1000{\text{ ml}}}} = \dfrac{1}{{50}}$

As both the carbonates are completely neutralised by ${\text{HCl}}$ so the relation between a number of g equivalent of ${\text{HCl}}$and both the carbonates is as follows:

${\text{Number of g equivalent of HCl}} = {\text{Number of g equivalent of CaC}}{{\text{O}}_{\text{3}}} + {\text{Number of g equivalent of N}}{{\text{a}}_{\text{2}}}{\text{CO}}$

$\dfrac{1}{{50}} = \dfrac{{{\text{x }}}}{{50}} + \dfrac{{(1.03 - x){\text{ }}}}{{53}}$

${\text{x = 0}}{\text{.5g}}$

As we assume the mass of ${\text{CaC}}{{\text{O}}_{\text{3}}}$ as x g

Hence, the mass of ${\text{CaC}}{{\text{O}}_{\text{3}}}$= 0.5g

Mass of ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$= (1.03-x) g = (1.03-0.5)g = 0.53g

Now using masses of each carbonate and mass of mixture we will calculate the percentage of sodium carbonate and calcium carbonate in the given mixture as follows:

{\text{ Mass % N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_3}{\text{ = }}\dfrac{{{\text{ Mass of N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_3}{\text{ }}}}{{{\text{ Mass of mixture }}}} \times 100\% = \dfrac{{0.53{\text{g}}}}{{1.03{\text{ g }}}} \times 100\% = 51.46\%

{\text{ Mass % CaC}}{{\text{O}}_{\text{3}}}{\text{ = }}\dfrac{{{\text{ Mass of CaC}}{{\text{O}}_{\text{3}}}{\text{ }}}}{{{\text{ Mass of mixture }}}} \times 100\% = \dfrac{{0.5{\text{g}}}}{{1.03{\text{ g }}}} \times 100\% = 48.54\%

Thus, the percentage of sodium carbonate and calcium carbonate in the given mixture is 51.46% and 48.54% respectively.

Note: The reaction between acid and base that give salt and water as the product is known as neutralisation reaction. When metal carbonates react with acid they give salt and carbonic acid. As carbonic acid is unstable it dissociates into water and carbon dioxide gas.