Question

An aqueous solution containing $6.5gm$ of $NaCl$ of $90\% $ purity was subjected to electrolysis. After the complete electrolysis, the solution was evaporated to get solid $NaOH$. The volume of $1M$ acetic acid required to neutralise NaOH obtained above is:
(A) $100c{m^3}$
(B) $200c{m^3}$
(C) $1000c{m^3}$
(D) $2000c{m^3}$

Answer 1

Hint: We know that sodium hydroxide is a strong acid. And acetic acid is also a weak acid. Salt and water are created when an acid and a base react. Neutralisation reaction is the name given to this phenomenon. The sodium hydroxide's molecular formula is $NaOH$. And acetic acid's molecular structure is $C{a_3}COOH$.

Complete Step by Step Solution:
In the question, an aqueous solution that contains $6.5gm$ of $NaCl$ with $90\% $ purity is given,

According to the given information, we notice that the weight of an aqueous solution is the multiple of $6.5gm$by $90\% $, so we have:
$Weight\,of\,pure\,NaCl = \dfrac{{6.5 \times 90}}{{100}} \\$
$\Rightarrow Weight\,of\,pure\,NaCl = 6.5 \times 0.9 \\$
$\Rightarrow Weight\,of\,pure\,NaCl = 5.85g \\$

As we know that an equivalent weight is obtained by dividing molecular weight by the valence of the element,
$Equivalent weight = \;\dfrac{{molecular mass}}{{valance of the element}}$
Molecular mass of $NaCl$ is $- \left( {22.98 amu of Na + 35.45 amu for Cl} \right) = 58.44$.

Since only one hydrogen atom could bond with $NaCl$ hence the valence is $1$.
So equivalent weight of $NaCl$ is $ = \dfrac{{5.85}}{{58.5}} \approx 0.1$
To determine the volume of $1M$acetic acid, multiply the equivalent weight of $NaCl$by $1M$, then we obtain:
$\dfrac{{0.1 \times 1000}}{1} = 100c{m^3}$
Therefore, the volume of $1M$ acetic acid required for the neutralisation of $NaOH$ is $100c{m^3}$.
Thus, the correct option is: (A) $100c{m^3}$.

Note: Always keep in mind that acetic acid and sodium hydroxide have a chemical equation that can be written as $NaOH + C{H_3}COOH \to {H_3}COONa + {H_2}O$. It is called the neutralisation reaction. Hydrolysis is the neutralisation reaction's opposite. The hydrolysis of a sodium acetate can be written as $C{H_3}COONa + {H_2}O \to C{H_3}COOH + NaOH$. Additionally, because sodium hydroxide is a potent base, the pH of the sodium hydroxide and acetic acid mixture will always be higher than $77$ . Therefore, the mixture will be basic in nature.