Question

A solution of 10 ml $\frac{M}{10}FeS{O}_4$ was titrated with $KMn{O}_4$ solution in acidic medium. The amount of $KMn{O}_4$ used will be:

• A. 5 ml of 0.1 M
• B. 10 ml of 0.1 M
• C. 10 ml of 0.5 M
• D. 10 ml of 0.02 M

Answer 1

The correct option is D

10 ml of 0.02 M

$2KMn{O}_4+3H_{2}S{O}_4\to K_{2}S{O}_4+2MnS{O}_4+3H_{2}O+5\left(O\right)$

$\frac{2FeS{O}_4+H_{2}S{O}_4+\left(O\right)\to F{e}_2(S{O}_4{)}_3+H_{2}O]\times 5}{2KMn{O}_4+8H_{2}S{O}_4+10FeS{O}_4\to Products}$

The n factor of iron sulphate is 1 here, meq of the salt solution: $10\times 0.1$ = 1 meq

The number of meq of iron sulphate will be the same as potassium permanganate. Remember, $N=M\times n$ where $N$ is the normality and $M$ is the molarity. Similarly, number of equivalents = number of moles $\times n$

The n factor of potassium permanganate is 5 in acidic medium as you can see. Among the given options, option D fits this:

$10\times 0.02\times 5$ = 1 meq