Question

0.66 gm of ${{H}_{3}}P{{O}_{2}}$ will require x ml of 0.1 M NaOH for complete neutralisation. x is
(A) 100 ml
(B) 200 ml
(C) 300 ml
(D) none of these

Answer 1

Hint: To solve the given illustration, we would require the basic knowledge of normality and neutralisation. The equations of neutralisation based on normality will help solve this problem more efficiently. Also, only one option can be correct here and before solving the problem we can ignore option (D) for a while.

Complete step by step solution:
Let us solve the given illustration directly, we just need to know the basics of normality as explained below:
Normality- Normality is the number of gram equivalents or mole equivalents of solute present in one litre of a solution. Mole equivalent is the number of moles who are actually the reactive units in the reaction. Similar goes with the gram equivalent.
It is expressed as,
\[N=M\times n\]
where, n is the ratio of molar mass to the equivalent mass of the component.
Normality can simply be expressed as,
\[N=\dfrac{{{n}_{g}}}{V}\]
where,
V = volume of solution in litres
${{n}_{g}}$ = number of gram equivalents i.e. it is the ratio of given mass to the equivalent mass of a compound.
-Illustration: Given that, Given mass of ${{H}_{3}}P{{O}_{2}}$ = 0.66 gm
Molarity of NaOH = 0.1 M
Molecular mass of ${{H}_{3}}P{{O}_{2}}$ = 66 gm/mol
Now, By the dissociation of ${{H}_{3}}P{{O}_{2}}$ and NaOH, we can say that,${{H}_{3}}P{{O}_{2}}$ has three ${{H}^{+}}$ and NaOH has one $O{{H}^{-}}$.
Thus,
Equivalent weight of ${{H}_{3}}P{{O}_{2}}$ = $\dfrac{66}{3}=22gmmo{{l}^{-1}}e{{q}^{-1}}$
And, number of gram equivalent is thus given as,
\[{{n}_{g}}\left( {{H}_{3}}P{{O}_{2}} \right)=\dfrac{0.66}{22}=0.03\]
Now, x moles of 0.1 M NaOH is required for complete neutralisation. Thus,
number of gram equivalent is given as,
\[{{n}_{g}}\left( NaOH \right)=\dfrac{xml\times 0.1M\times 1}{1000L}=\dfrac{x}{10000}\]
So, at complete neutralisation,
\[\begin{align}
& {{n}_{g}}\left( {{H}_{3}}P{{O}_{2}} \right)={{n}_{g}}\left( NaOH \right) \\
& 0.03=\dfrac{x}{10000} \\
& \therefore x=300ml \\
\end{align}\]
Thus, 300 ml of NaOH is required for the process.

Therefore, option (C) is correct.

Note: Do note to use the units properly and do not get confused in concepts of molarity and normality. Though they are interrelated, their multiplying factors differ.