Question

$0.1M$ $KMn{O_4}$ is used for the following titration. What volume of the solution in $mL$ will be required to react with $0.158g$ of $N{a_2}{S_2}{O_3}$?
${S_2}O_3^{2 - } + MnO_4^ - + {H_2}O\xrightarrow{{}}Mn{O_2}\left( s \right) + SO_4^{2 - } + O{H^ - }$

Answer 1

Hint: Titration is the common laboratory method of quantitative chemical analysis to determine the concentration of an identified analyte. Molarity is used to define concentration of a solution. Molarity is defined as the number of moles that are present per liter of the solution.
Formula used: Normality$ = $molarity$ \times $oxidation state
Number of equivalents$ = $ moles$ \times $oxidation state
Number of moles$ = \dfrac{{{\text{given mass }}}}{{{\text{molecular mass}}}}$
${N_1}{V_1} = {N_2}{V_2}$
Where ${N_1}$ and ${V_1}$ is normality and molarity of one substance and ${N_2}$ and ${V_2}$ is normality and molarity of other substances.
normality$ \times $ volume$ = NV = $ number of equivalents

Complete step by step answer:
$KMn{O_4}$ is potassium permanganate. It acts as a self indicator. It means that this is a chemical substance which can mark the end point of the titration or any other reaction along with self participation in the reaction.
In given reaction $KMn{O_4}$ changes are:
$MnO_4^ - \xrightarrow[{}]{}Mn{O_2}$
Oxidation state of $Mn$ in $KMn{O_4}$ on reactant side is $7$ (charge on complete $MnO_4^ - $ ion is $ - 1$ and on four atoms of oxygen is $ + 8$ so oxidation state of $Mn$ is $7$). Oxidation state of $Mn$ on the product side is $ + 4$ (compound is neutral and charge on two oxygen atoms is $ - 4$ so charge on $Mn$ is $ + 4$). Change in oxidation state of $Mn$ is $7 - 4 = 3$.
Molarity of $KMn{O_4}$ is $0.1M$ (given) therefore normality of $KMn{O_4}$ is (using above formula)
 Normality$ = $molarity$ \times $oxidation state
Normality$ = 0.1 \times 3$ (molarity is $0.1M$ and oxidation state is $3$)
Normality$ = 0.3$
Similarly oxidation state of two atoms of sulphur on reactant side is $ + 4$ (as charge on molecule ${S_2}O_3^{2 - }$ is $ - 2$ and charge on three oxygen atoms is $ - 6$). Oxidation state of two atoms on the sulphur on the product side is $ + 12$ (oxidation state on the whole compound is an oxidation state of oxygen atoms is $ - 8$). We have taken two atoms because there are two sulphur atoms on the left side. So the change in oxidation state of sulphur is $12 - 4 = 8$.
Molecular mass of $N{a_2}{S_2}{O_3}$ is $158g$ (calculated by calculating mass of all the atoms of the compound)
Given mass of $N{a_2}{S_2}{O_3}$ is $0.158g$
Therefore number of moles of $N{a_2}{S_2}{O_3}$ is:
Number of moles$ = \dfrac{{{\text{given mass }}}}{{{\text{molecular mass}}}}$
Number of moles$ = \dfrac{{0.158}}{{158}}$
Number of moles$ = 1 \times {10^{ - 3}}$
And
Number of equivalents$ = $ moles$ \times $oxidation state
Moles are $1 \times {10^{ - 3}}$ and oxidation state is $8$
So, Number of equivalents$ = $$1 \times {10^{ - 3}} \times 8$
Number of equivalents$ = $$8 \times {10^{ - 3}}$
Equation of titration is ${N_1}{V_1} = {N_2}{V_2}$
Let volume of $KMn{O_4}$ be $V$ liter and equal to ${V_1}$, ${N_1}$ be normality of $KMn{O_4}$ which is equal to $0.3$ (calculated above)
As we know $NV = $ number of equivalents
Number of equivalents of $N{a_2}{S_2}{O_3}$ is $8 \times {10^{ - 3}}$ which is equal to ${N_2}{V_2}$. So, applying the equation:
${N_1}{V_1} = {N_2}{V_2}$
$0.3 \times \dfrac{V}{{1000}} = 8 \times {10^{ - 3}}$ ($\dfrac{V}{{1000}}$ is taken because we need volume in $mL$)
Solving this we get, $V = 26.67mL$
So, the volume of $KMn{O_4}$ is $26.67mL$.

Note:
Concentration of a substance is inversely proportional to the volume of solution. According to the osmotic pressure equation $\Pi = iCRT$ where $C$ is concentration and $T$ is temperature, we can conclude that concentration is inversely proportional to the temperature i.e.
$C \propto \dfrac{1}{T}$
And concentration is inversely proportional to volume so volume is directly proportional to temperature i.e.
$V \propto T$