Question

If x % of oxalate ion in a given sample of oxalate salt of which $\text{0}\text{.6g}$ dissolved in $\text{100mL}$ of water required $\text{90mL}$ of $\dfrac{\text{M}}{\text{100}}\text{KMn}{{\text{O}}_{\text{4}}}$ for complete oxidation, then the value of x is
A) 33
B) 40
C) 30
D) None of these

Answer 1

Hint: Redox titrations also called an oxidation – reduction titration can accurately determine the concentration of an unknown analyte by measuring it against a standardized titrant. Titration of Potassium Permanganate $\left( KMn{{O}_{4}} \right)$ against oxalic acid is an example of redox titration. At equivalence point the milliequivalents of oxalate ion equals the milliequivalents of $MnO_{4}^{2-}$ .

Complete step by step answer:
Given, 0.6 g of 100mL of oxalate ion required 90mL of $\dfrac{M}{100}KMn{{O}_{4}}$ for complete oxidation.
We know that reaction of Potassium Permanganate with oxalic acid is a type of redox titration where $KMn{{O}_{4}}$ acts as oxidizing agent and oxalic acid acts as reducing agent.
Redox changes occurring in the reaction –
$MnO_{4}^{-}+8{{H}^{+}}+5{{e}^{-}}\to M{{n}^{2+}}+4{{H}_{2}}O$
${{H}_{2}}{{C}_{2}}{{O}_{4}}\to 2C{{O}_{2}}+2{{e}^{-}}+2{{H}^{+}}$
So overall reaction
$2MnO_{4}^{-}+5{{H}_{2}}{{C}_{2}}{{O}_{4}}+6{{H}^{+}}\to 2M{{n}^{2+}}+8{{H}_{2}}O$
Therefore,
Milliequivalent of oxalate ion = Milliequivalent of $MnO_4^-$
Now,
Equivalent of oxalate ion = ${\dfrac{Weight\;of \;oxalate\; ion}{Equivalent\; weight\; of\; \;oxalate\; ion}}$
Since, Equivalent weight of $(C_{2}O_{4})^{2-}$ ion = $\dfrac{Molecular\; weight}{n-factor}$
= $\dfrac{88}{2}$
= 44
Let W be the weight of oxalate ions.
Milliequivalent of oxalate ion = Equivalent of oxalate $\times {{10}^{3}}$
\[=\dfrac{w}{Molecular\; weight}\times n-factor\times {{10}^{3}}\]
\[=\dfrac{w}{Equivalent\;weight}\times {{10}^{3}}\]
\[=\dfrac{w}{44}\times {{10}^{3}}\]
Similarly, $Milliequivalent\text{ of MnO}_{4}^{-}$ = \[\dfrac{Weight}{Molar\; weight}\times n-factor\times {{10}^{3}}\]
\[\because\dfrac{Weight}{Molar\; weight}=No.\;of\; moles\]
And
No. of moles = Concentration × Volume
\[=\dfrac{1}{100}\times 90\]
$Milliequivalent\; of \;MnO_{4}^{-}$ = $90\times \dfrac{1}{100}\times 5$
Therefore, ${\dfrac{W\times 10^{3}}{44}}$ = $90\times \dfrac{1}{100}\times 5$
Solving the equation we get –
$Weight\; of\; {C}_{2}O_{4}^{2-}$ = 0.198 g
Hence, \[x%\text{ of }0.6\text{g of oxalate ion = }0.198\text{g}\]
$\Rightarrow \dfrac{x}{100}\times 0. 6=0.198$
$\Rightarrow x=\dfrac{0.198\times 100}{0.6}$
Solving the above equation, we get x=33 % .

Hence option (A) is the correct option.

Note: Students must be thorough with the concept of calculation of n – factor. Oxalic acid which is a dicarboxylic acid has a basicity equals to 2 so \[{{n}_{f}}\] for oxalic acid is 2 while for $MnO_{4}^{-}$ ion ${{n}_{f}}=5$ since it reduces itself to form $M{{n}^{+2}}$ losing 5 electrons.