Question

$0.124g$ of iron wire was dissolved in dilute ${H_2}S{O_4}$ in oxygen free atmosphere and the resultant solution was titrated against $0.09672N$ solution of $KMn{O_4}$. The titre value was $22.90mL$. Calculate the percentage purity of iron wire

Answer 1

Titration is a process in which there is a slow addition of a titrant into a titre in order to attain neutralization. The point at which there is a change in color of the indicator when the titrant is added in the titer slowly, it is marked as the neutralization point

Complete answer:
When iron wire is dissolved in dilute sulfuric acid, the following reaction takes place:
$Fe(s) + {H_2}S{O_4}(aq) \to FeS{O_4}(aq) + {H_2}(g)$
In this reaction, the iron replaces the hydrogen from sulfuric acid and forms an aqueous solution of ferrous sulfate.
Now, according to the question, the equivalent of the potassium permanganate is neutralized by the equivalents of ferrous sulfate.
$22.90mL$ of $0.09672N$ $FeS{O_4}$ $\equiv$ $22.90mL$ of $0.09672N$ $KMn{O_4}$
Now, we need to determine the amount of $FeS{O_4}$ in the solution.
The molecular weight of $FeS{O_4}$ is = $56 + 32 + 16 \times 4 = 152g$
Amount of $FeS{O_4}$ = $\dfrac{{0.09672 \times 152 \times 22.90}}{{1000}} = 0.3366g$
Now, we need to find out the amount of iron in $0.3366g$ of $FeS{O_4}$.
Thus, the amount of iron in $0.3366g$ of $FeS{O_4}$ = $\dfrac{{56}}{{152}} \times 0.3366 = 0.124g$
In order to find the percentage purity of iron wire, we need to apply the mathematical relation:
$\% purity = \dfrac{{{w_{cal}}}}{{{w_{theo.}}}} \times 100$
Now, substituting the values and on solving, we have:
$\% purity = \dfrac{{0.124}}{{0.124}} \times 100 = 100\%$
Hence, the iron wire is $100\%$ pure in the given question.

Note:
The ferrous sulfate is used in the medicines for the treatment of anaemia. The iron present in the medicines help the body to produce the red blood cells which are the career of oxygen in our body. Thus, iron sulfate can act as a good supplement of iron in our body when taken orally.