Question

What is the molarity of a solution of nitric acid if $0.216$ of barium hydroxide is required to neutralize $20mL$ of nitric acid?

Answer 1

Hint : Molarity is a term which is used to calculate the amount of the substance present in a particular volume of a solution. For that we will find out the number of moles first and then using them we can find out the molarity.

Complete Step By Step Answer:
First of all, we will write down its chemical reaction:
$Ba{(OH)_{2(aq)}} + 2HN{O_3} \to Ba{(N{O_3})_2}_{(aq)} + 2{H_2}{O_{(l)}}$
Here, $aq$and $l$ are the symbols used to express the state of the compounds. $l$ and $aq$ stands for liquid and aqueous state.
Now, we will find out the number of moles of $Ba{(OH)_2}$
$n = \dfrac{m}{M}$, where n, m and M stands for number of moles, mass and the molecular mass
Therefore, Molecular mass of $Ba{(OH)_2}$$$$ = 137 + 2(16 + 1) = 171gmo{l^{ - 1}}$
Given mass is $0.216$
After putting the values in the formula:
$n = \dfrac{{0.216 \cdot g}}{{171 \cdot g \cdot mo{l^{ - 1}}}} = 1.26 \times {10^{ - 3}}mol$
Now, as we can see from the chemical equation there is $1:2$ratio of both the reactants. So, we can find out the number of moles for $HN{O_3}$ by multiplying it by its stoichiometric coefficient:
$n = 2.52 \times {10^{ - 3}}mol$ of nitric acid.
Now, let’s find out the molarity of nitric acid using its formula:
$M = \dfrac{n}{V}$, where n and V stands for the number of moles of solute and volume of solution in litres
So, the concentration of nitric acid $ = \dfrac{n}{V} = \dfrac{{2.52 \times {{10}^{ - 3}}mol}}{{20 \times {{10}^{ - 3}} \cdot L}} \simeq 0.126 \cdot mol \cdot {L^{ - 1}}$
Therefore, $0.126 \cdot mol \cdot {L^{ - 1}}$ solution of nitric acid is required to neutralize $20mL$of nitric acid.

Note :
As we have seen that such type of numerical are easy to solve but sometimes we face difficulties in the calculations due to the difference in units, so, we have to take care of units as well, we need to use all the units in the same system.