Question

$1.5g$ of sample of impure potassium dichromate was dissolved in water and made up to $500mL$ solution. $25mL$ of this solution required iodometrically $24mL$ of a sodium thiosulphate solution. $26mL$ of this sodium thiosulphate solution required $25mL$ of $N/20$ solution of pure potassium dichromate. Find the percentage purity of the impure sample of potassium dichromate.

Answer 1

Normality of sodium thiosulphate solution any be determined as:
$N_1{V}_1\left(N{a}_2{S}_2{O}_4\right)=N_2{V}_2\left(pure{K}_{2}C{r}_2{O}_7\right)$
$N_1\times 26=25\times \frac{1}{20}$
$N_1=0.048\left(hypo\right)$
The reaction involved may be given as:
$C{r}_2{O}_7^{2-}+6I^{-}+4H^{+}\to 2C{r}^{3+}+3I_2+7H_{2}O$
$3[I_2+2N{a}_2{S}_2{O}_3\to 2NaI+N{a}_2{S}_4{O}_6]$
$1mole{K}_{2}C{r}_2{O}_7\equiv 6moleN{a}_2{S}_2{O}_3....\left(i\right)$
$25mL$ of solution of $K_{2}C{r}_2{O}_7$ is treated by $24mL$ of $0.048N$ hypo
$\therefore 500mL$ of solution will be titrated by $480L$ of $0.048N$ hypo
No. of moles of hypo $=\frac{Mass}{M.w.\left(158\right)}=\frac{E\times N\times V}{1000\times 158}$
$=\frac{158\times 0.048\times 480}{1000\times 158}$
$=0.02304mole$
No. of moles of $K_{2}C{r}_2{O}_7=\frac{1}{6}$ [No. of moles of hypo]
$=\frac{1}{6}\left[0.02304\right]=3.84\times {10}^{-3}$
Mass of $K_{2}C{r}_2{O}_7=3.84\times {10}^{-3}\times 294=1.12896$
% $purity=\frac{1.12896}{1.5}\times 100=75.26$%.