Question

How many milliliters of \[0.5{\text{M}}\] \[{{\text{H}}_2}{\text{S}}{{\text{O}}_4}\] are needed to dissolve \[0.5{\text{g}}\] of copper (II) carbonate?

Answer 1

Hint: We should have a basic knowledge of the chemical reaction that takes place.
The formula used for the calculation of molarity is shown below:
\[{\text{M}} = \dfrac{{\text{n}}}{{{\text{V}}\left( {\text{L}} \right)}}\]
Where M is molarity
n is number of moles
V is volume of solution in liters
Number of moles can be calculated by using the formula:
\[{\text{n = }}\dfrac{{\text{m}}}{{\text{M}}}\]
Where n is number of moles
m is given mass
M is molar mass

Complete answer:
First step is to write the reaction that is involved:
When copper carbonate is treated with sulphuric acid, the products formed are copper sulphate, carbon dioxide and water molecules.
The complete balanced reaction is shown below:
\[{\text{CuC}}{{\text{O}}_3} + {{\text{H}}_2}{\text{S}}{{\text{O}}_4} \to {\text{CuS}}{{\text{O}}_4} + {\text{C}}{{\text{O}}_2} + {{\text{H}}_2}{\text{O}}\]
Molar mass of the compound can be defined as the sum of the atomic masses of all the elements present in that compound. Its unit is \[{\text{g/mol}}{\text{.}}\]
Now calculate the molar masses of copper carbonate and sulphuric acid.
Molar mass of
\[{\text{CuC}}{{\text{O}}_3} = 63.5 + 12 + \left( {16 \times 3} \right)\]
On simplification we get,
\[{\text{CuC}}{{\text{O}}_3} = 123.5{\text{g/mol}}\]
Now, for sulphuric acid:
\[{{\text{H}}_2}{\text{S}}{{\text{O}}_4} = \left( {2 \times 1} \right) + 32 + \left( {16 \times 4} \right)\]
On simplification we get,
\[{{\text{H}}_2}{\text{S}}{{\text{O}}_4} = 98{\text{g/mol}}\]
Now,
\[123.5{\text{gofCuC}}{{\text{O}}_3} \to 98{\text{gof}}{{\text{H}}_2}{\text{S}}{{\text{O}}_4}\]
\[0.5{\text{gofCuC}}{{\text{O}}_3} \to ?\]
\[ = \dfrac{{0.5{\text{gofCuC}}{{\text{O}}_3} \times 98{\text{gof}}{{\text{H}}_2}{\text{S}}{{\text{O}}_4}}}{{123.5{\text{gofCuC}}{{\text{O}}_3}}}\]
On simplification we get,
$0.397g$ of ${H_2}S{O_4}$
Now, next step is to calculate the number of moles as shown below:
$n = \dfrac{m}{M}$
Now we can substitute the known values we get,
$n = \dfrac{{0.397}}{{98}}$
On simplification we get,
$n = 0.0040moles$
Now, the last step is to calculate the molarity,
\[{\text{M}} = \dfrac{{\text{n}}}{{{\text{V}}\left( {\text{L}} \right)}}\]
Now we can substitute the known values we get,
\[{\text{V}}\left( {\text{L}} \right) = \dfrac{{0.0040}}{{0.5}}\]
On simplification we get,
\[{\text{V}}\left( {\text{L}} \right) = 0.0081{\text{L}}\]
Now we convert liter into milliliter we get,
\[{\text{V}}\left( {{\text{ml}}} \right) = 8.1{\text{ml}}\]
Hence, the volume required in milliliters is \[8.1{\text{ml}}\].

Note:
Students might make a mistake with unit conversion. It is always better to use dimensional analysis while solving any numerical problem.
Please take care to convert the volume in liters to ml, as in the question it is asked in ml and not in L.
Conversion factor: \[1{\text{L = 1000ml}}\]
Molarity is defined as the ratio of the number of moles of solute to that of volume of solution in liters.