Question

What volume of $0.25$ M $HN{O_3}$ (nitric acid) solution reacts with $50$ ml of $0.15$ M $N{a_2}C{O_3}$ (sodium carbonate) solution in the following reaction?
$2HN{O_{3\left( {aq} \right)}} + N{a_2}C{O_{3\left( {aq} \right)}} \to 2NaN{O_{3\left( {aq} \right)}} + {H_2}{O_{\left( l \right)}} + C{O_{2\left( g \right)}}$
(A) $6$ ml
(B) $60$ ml
(C) $30$ ml
(D) None

Answer 1

Nitric acid is a strong acid and sodium carbonate is a base, when a strong acid and a base react with each other, leads to the formation of a salt along with the formation of water and the liberation of a carbon dioxide gas. The salt formed will be sodium nitrate, which is an ionic or inorganic salt.

Complete answer:
Given that nitric acid is reacted with sodium carbonate.
Nitric acid is a strong acid with the molecular formula of $HN{O_3}$
Sodium carbonate is a base with molecular formula of $N{a_2}C{O_3}$
The molarity of nitric acid is $0.25$ M
The volume of nitric acid has to be determined
The molarity of sodium carbonate is $0.15$ M
The volume of sodium carbonate is $50$ ml
In this reaction two moles of nitric acid were equivalent to one mole of sodium carbonate.
The number of moles of sodium carbonate will be $0.15 \times 0.05 = 0.0075mol$
The number of moles of nitric acid will be $0.0075 \times 2 = 0.015mol$
Thus, the volume will be equal to $\dfrac{{0.015}}{{0.25}} = 0.06L$
When converting these litres to millilitres we will get $60$ ml.
Thus, option $2$ is the correct one.

Note:
Molarity or molar concentration can be obtained by dividing the number of moles with volume of solution in litres. Firstly, we find out the moles of sodium carbonate. With the help of moles of sodium carbonate the moles of sodium nitric acid is calculated. Finally, volume is calculated from molarity and number of moles of nitric acid.