Question

Given,$POC{l_3}$ hydrolysed in water to give ${H_3}P{O_4}$ and ${\text{HCl}}$ only. What volume of $2{\text{ M Ca}}{\left( {OH} \right)_2}$ is needed to completely react with $50{\text{ ml , 0}}{\text{.1 M}}$ $POC{l_3}$ solution.
$(i){\text{ 5 ml}}$
$(ii){\text{ 7}}{\text{.5 ml}}$
$(iii){\text{ 15 ml}}$
$(iv){\text{ 10 ml}}$

Answer 1

Here we need to calculate the volume of $2{\text{ M Ca}}{\left( {OH} \right)_2}$ which will completely react with $50{\text{ ml , 0}}{\text{.1 M}}$ $POC{l_3}$solution. We will find the number of equivalents of the ${H_3}P{O_4}$ and ${\text{HCl}}$,then by using normality equation, we will find the volume of $2{\text{ M Ca}}{\left( {OH} \right)_2}$.
Formula Used:
${{\text{N}}_1}{{\text{V}}_1}{\text{ = }}{{\text{N}}_2}{{\text{V}}_2}$

Complete Answer:
When we add water to $POC{l_3}$, we get the Phosphoric acid and hydrochloric acid. Phosphoric acid is a weak acid while hydrochloric acid is strong acid. The reaction between water and $POC{l_3}$ can be represented as,
${{\text{V}}_2}{\text{ = 7}}{\text{.5 ml}}$
Thus three moles of water is consumed with $POC{l_3}$ to produce one mole of ${{\text{H}}_3}{\text{P}}{{\text{O}}_4}$ and three moles of ${\text{HCl}}$. Now we will find the equivalents of both ${H_3}P{O_4}$ and ${\text{HCl}}$. The number of equivalent can be calculated as-
${\text{equivalent = volume }} \times {\text{ molarity }} \times {\text{ n - factor}}$
Thus for ${H_3}P{O_4}$, number of equivalent is,
Volume of solution $= {\text{ }}50{\text{ ml}}$
Molarity of solution $= {\text{ 0}}{\text{.1 M}}$
${\text{n - factor = 3}}$
${\text{equivalent = 50 }} \times {\text{ 0}}{\text{.1 }} \times {\text{ 3}}$
Similarly we can calculate the equivalent of ${\text{HCl}}$produced. Since three moles of ${\text{HCl}}$ we have to multiply the equivalent by three. Thus for, ${\text{HCl}}$ number of equivalent is,
${\text{equivalent = 50 }} \times {\text{ 0}}{\text{.1 }} \times {\text{ 1 }} \times {\text{ 3}}$
We know that the product of normality and volume of solution is equal to equivalents of solute. Also Therefore,
${{\text{N}}_{\text{1}}}{{\text{V}}_1} = {\text{ 6 }} \times {\text{ 50 }} \times {\text{ 0}}{\text{.1}}$.
Let the volume of $Ca{\left( {OH} \right)_2}$ be ${{\text{V}}_2}$. Hence we can find the normality ${N_2}$ of $Ca{\left( {OH} \right)_2}$ as .
${N_2} = {\text{ Molarity }} \times {\text{ n - factor}}$
${N_2} = {\text{ 2 }} \times {\text{ 2}}$
Now we know that from equation of normality we have,
${{\text{N}}_1}{{\text{V}}_1}{\text{ = }}{{\text{N}}_2}{{\text{V}}_2}$
Substituting the values we get,
${\text{ 6 }} \times {\text{ 50 }} \times {\text{ 0}}{\text{.1 = 2 }} \times {\text{ 2 }} \times {\text{ }}{{\text{V}}_2}$
On solving the equation we get,
${{\text{V}}_2}{\text{ = 7}}{\text{.5 ml}}$
Thus the volume of $Ca{\left( {OH} \right)_2}$ required is ${\text{7}}{\text{.5 ml}}$.

Note:
The product of normality and the volume of the solution gives us the number of the equivalents of the solute. Here the equivalent is called milli-equivalents because the volume is in milli-litre. Also n-factor is calculated as the basicity for base and acidity for an acid. Since Phosphoric acid can donate three hydrogen ions to a base therefore its n-factor is three.