Question

How many ml of a .10M NaOH solution are needed to neutralize 15ml of .20M ${{H}_{3}}P{{O}_{4}}$ solution?

Answer 1

This question can be formed by using the normality of the chemical compounds. Normality (N) is a method of measurement of concentration of a solution and is expressed in terms of mole equivalents of the solute per liter of solution. The SI unit for morality is eq/L.

Complete step-by-step answer: The balanced equation of the neutralization reaction between NaOH and ${{H}_{3}}P{{O}_{4}}$ is as follows
$3NaOH(aq)+{{H}_{3}}P{{O}_{4}}(aq)\to N{{a}_{3}}P{{O}_{4}}+3{{H}_{2}}O$
As we can see that 3 moles of NaOH are required to neutralize 1 mole of ${{H}_{3}}P{{O}_{4}}$.
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.
Molarity can be converted into normality by using the following 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 NaOH contributes just one $O{{H}^{-}}$ ion in the reaction, hence the number of equivalents n = 1. So, the normality of the 0.1M NaOH will be
$N=0.1\times 1=0.1N$
Also, we can see ${{H}_{3}}P{{O}_{4}}$ that contributes three ${{H}^{+}}$ ion in the reaction, hence the number of equivalents n = 3. So, the normality of the 0.2M ${{H}_{3}}P{{O}_{4}}$will be
$N=0.2\times 3=0.6N$
Now, in neutralization reactions, the relationship between the volume and normality of an acid and a base is given by the formula
$(acid)\text{ }{{N}_{1}}{{V}_{1}}={{N}_{2}}{{V}_{2}}\text{ (base)}$
Since we already know that ${{N}_{1}}=0.6N,\text{ }{{V}_{1}}=15ml,\text{ }{{N}_{2}}=0.1N$, ${{V}_{2}}$ will be
${{V}_{2}}=\dfrac{0.6\times 15}{0.1}=90ml$
So, 90ml of 0.10M NaOH is required to neutralize 15ml of 0.20M ${{H}_{3}}P{{O}_{4}}$ 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 ${{H}_{3}}P{{O}_{4}}$ is a polyprotic acid and donates more than one ${{H}^{+}}$ ion in the reaction and hence the mole ratio is not 1:1.