Question

How many mL of 0.200 M HCl are needed to neutralize 20.0 ml of 0.150 M $Ba{{(OH)}_{2}}$?

Answer 1

The concept of normality of chemical solutions can be used to solve this question. One of the methods of measurement of concentration of a solution is normality (N) and is expressed in terms of mole equivalents of the solute per liter of solution. The SI unit for morality is eq/L.

Complete answer:
The balanced chemical equation of the neutralization reaction between HCl and $Ba{{(OH)}_{2}}$ is as follows
$2HCl+Ba{{(OH)}_{2}}\to BaC{{l}_{2}}+2{{H}_{2}}O$
As we can see that 2 moles of HCl are required to neutralize 1 mole of $Ba{{(OH)}_{2}}$.
Now we know that when the concentration of a solute in a solution is measured in terms of amount of solute per unit volume, we get the molarity M of the solution, or the molar concentration. The SI unit for molarity is mol/L.
Normality of a solution can be obtained by its molarity using the formula
$N=M\times n$
Where, n is the number of equivalents can be given by the number of ${{H}^{+}}$ ions or $O{{H}^{-}}$ ions which are donated by an acid or a base in a given reaction.
In the given reaction, we can see that HCl contributes just one ${{H}^{+}}$ ion in the reaction, hence the number of equivalents n=1. So, the normality of 0.200M HCl solution will be
$N=0.200\times 1=0.2N$
Also, we can see that $Ba{{(OH)}_{2}}$ contributes two $O{{H}^{-}}$ ions in the reaction, hence the number of equivalents n=2. So, the normality of the 0.150M $Ba{{(OH)}_{2}}$ solution will be
$N=0.15\times 2=0.3N$
The relationship between the normality of the solution and the volume of the solution of two reactants during a neutralization reaction is given by the formula
$(acid)\text{ }{{N}_{1}}{{V}_{1}}={{N}_{2}}{{V}_{2}}\text{ (base)}$
Since we already know that ${{N}_{1}}=0.2N,\text{ }{{V}_{2}}=20ml,\text{ }{{N}_{2}}=0.3N$, ${{V}_{1}}$ will be
${{V}_{1}}=\dfrac{0.3\times 20}{0.2}=30ml$
So, 30ml of 0.200M HCl solution is required to neutralize 20ml of 0.150M $Ba{{(OH)}_{2}}$ solution.

Note:
It is important to note that we cannot use the molarity neutralization equation ${{M}_{1}}{{V}_{1}}={{M}_{2}}{{V}_{2}}$ directly since $Ba{{(OH)}_{2}}$ is a polyprotic base and donates more than one $O{{H}^{-}}$ ion in the reaction and hence the mole ratio is not 1:1.