Question

When $10$mL ${\text{NaH}}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}$is oxidised by $10$mL of $0.02$ M ${\text{MnO}}_4^ - $. Hence 10 mL of ${\text{NaH}}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}$is neutralised by:
A. $10$mL of$0.1$M ${\text{NaOH}}$
B. $10$mL of$0.02$M ${\text{NaOH}}$
C. $10$mL of$0.1$M ${\text{Ca(OH}}{{\text{)}}_2}$
D. $10$mL of$0.05$M ${\text{Ba(OH}}{{\text{)}}_2}$

Answer 1

Hint: The first reaction of ${\text{NaH}}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}$ with ${\text{MnO}}_4^ - $ is the redox reaction. From this reaction we will determine the valance factor for ${\text{MnO}}_4^ - $ so, we can determine its normality. Then by using normality and volume relation for titration we can determine the normality of ${\text{NaH}}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}$. Now, we know the in neutralization reaction normality and volume product for both the titrant and titrate will remain the same. So, after determining the normality of acid ${\text{NaH}}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}$, we will choose the base from the given option whose normality is same as for ${\text{NaH}}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}$.
Formula used: ${{\text{N}}_{\text{1}}}{{\text{V}}_{\text{1}}}\,{\text{ = }}\,{{\text{N}}_{\text{2}}}{{\text{V}}_{\text{2}}}$
Normality = ${\text{molarity}}\,\,{\text{ \times }}\,{\text{valence factor}}$

Complete step-by-step answer:
The reaction of ${\text{NaH}}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}$with ${\text{MnO}}_4^ - $ is the redox reaction.
In this redox reaction, ${\text{MnO}}_4^ - $ is an oxidizing agent.${{\text{C}}_{\text{2}}}{\text{O}}_4^{2 - }$ oxidised to ${\text{C}}{{\text{O}}_{\text{2}}}$ and ${\text{MnO}}_4^ - $ get reduce to ${\text{M}}{{\text{n}}^{{\text{2 + }}}}$.
Oxidation: \[\mathop {{{\text{C}}_{\text{2}}}}\limits_{ + 3} {\text{O}}_4^{2 - }\,\,\, \to \mathop {\text{C}}\limits_{ + 4} {{\text{O}}_{\text{2}}} + \,2{{\text{e}}^ - }\]
Reduction: $\mathop {{\text{Mn}}}\limits_{ + 7} {\text{O}}_4^ - \, + \,5{{\text{e}}^ - }\,\, \to \mathop {{\text{M}}{{\text{n}}^{{\text{2 + }}}}}\limits_{ + 5} $
During the reaction ${\text{MnO}}_4^ - $ is accepting five electrons, so the valence factor of ${\text{MnO}}_4^ - $ is five.
Normality is determined by multiplying the molarity with valance factor.
Normality = ${\text{molarity}}\,\,{\text{ \times }}\,{\text{valence factor}}$
So, normality of ${\text{MnO}}_4^ - $ = ${\text{molarity}}\,\,{\text{ \times }}\,\,{\text{5}}$
On substituting $0.02$M for molarity of ${\text{MnO}}_4^ - $,
Normality of ${\text{MnO}}_4^ - $ = ${\text{0}}{\text{.02}}\,\,{\text{ \times }}\,\,{\text{5}}$
Normality of ${\text{MnO}}_4^ - $ = ${\text{0}}{\text{.1}}$N
Now, for complete the redox reaction, equivalent amount of both should be same so,
${{\text{N}}_{\text{1}}}{{\text{V}}_{\text{1}}}\,{\text{ = }}\,{{\text{N}}_{\text{2}}}{{\text{V}}_{\text{2}}}$
Where,
${{\text{N}}_{\text{1}}}$is the normality of the solution having ${{\text{V}}_{\text{1}}}$volume.
${{\text{N}}_{\text{2}}}$is the normality of the solution having ${{\text{V}}_{\text{2}}}$volume.
We assume that ${{\text{N}}_{\text{1}}}$ is the normality of ${{\text{C}}_{\text{2}}}{\text{O}}_4^{2 - }$ and ${{\text{N}}_{\text{2}}}$ is he normality of ${\text{MnO}}_4^ - $.
On substituting, $10$mL for ${{\text{V}}_{\text{1}}}$, ${\text{0}}{\text{.1}}$N for ${{\text{N}}_{\text{1}}}$ and$10$ mL for${{\text{V}}_{\text{2}}}$,
${{\text{N}}_{\text{1}}} \times \,10\,\,{\text{ = }}\,{\text{0}}{\text{.1}} \times \,10$
${{\text{N}}_{\text{1}}}{\text{ = }}\,{\text{0}}{\text{.1}}\,{\text{N}}$
So, the normality of the ${{\text{C}}_{\text{2}}}{\text{O}}_4^{2 - }$ is ${\text{0}}{\text{.1}}$N.
In neutralization reaction one equivalent of the acid completely reacts with one equivalent of base so the normality of the base will also be ${\text{0}}{\text{.1}}$N.
The base given option A is, $10$mL of$0.1$M ${\text{NaOH}}$.
The neutralization reaction of ${\text{NaH}}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}$ with NaOH is as follows:
\[{\text{NaH}}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}\, + \,{\text{NaOH}}\, \to \,{\text{N}}{{\text{a}}_2}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}\, + \,{{\text{H}}_2}{\text{O}}\]
The NaOH dissociates as, ${\text{NaOH}}\, \to \,{\text{N}}{{\text{a}}^ + }\, + \,{\text{O}}{{\text{H}}^ - }$
Due to exchange of one electron the valence factor for NaOH is one.
So, normality of NaOH = ${\text{molarity}}\,{\text{ \times 1}}$
Normality of NaOH = ${\text{0}}{\text{.1 \times 1}}$
Normality of NaOH = ${\text{0}}{\text{.1}}$N
So, when neutralization reaction take place, $10$ mL of ${\text{0}}{\text{.1}}$ N ${{\text{C}}_{\text{2}}}{\text{O}}_4^{2 - }$ will completely neutralised by $10$ mL of ${\text{0}}{\text{.1}}$ N NaOH.
We can check it as,
${{\text{N}}_{{{\text{C}}_{\text{2}}}{\text{O}}_4^{2 - }}}{{\text{N}}_{{{\text{C}}_{\text{2}}}{\text{O}}_4^{2 - }}}\,{\text{ = }}\,{{\text{N}}_{{\text{NaOH}}}}{{\text{V}}_{{\text{NaOH}}}}$
${\text{0}}{\text{.1}}\, \times \,{\text{10}}\,{\text{ = }}\,{\text{0}}{\text{.1}}\, \times \,{\text{10}}$
So, 10 mL of ${\text{NaH}}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}$is neutralised by $10$mL of$0.1$M ${\text{NaOH}}$.

Therefore, option (A) $10$mL of$0.1$M ${\text{NaOH}}$ is correct.

Note: During acid-base titration neutralization reaction takes place. In this reaction known volume and unknown concentration of acid is titrated with a base. So, the concentration of the unknown is determined. Valency factor is the number of electrons gained or losses during a reaction for the proton donated. The valency factor for T ${\text{NaH}}{{\text{C}}_{\text{2}}}{{\text{O}}_{\text{4}}}$ in redox reaction is $2$but the valency factor in neutralization reaction is \[1\].