Question

$5g$ of a sample of brass were dissolved in $1litre$ conc. $H_{2}S{O}_4.20mL$ of this solution were mixed with $KI$ and liberated iodine required $20mL$ of $0.0327N$ hypo solution for titration. Calculate the amount of copper in the alloy.

Answer 1

When brass is extracted with concentrated $H_{2}S{O}_4$, it gives copper sulphate.
$2[Cu+2H_{2}S{O}_4\to CuS{O}_4+S{O}_2+2H_{2}O]$
$2CuS{O}_4+4KI\to 2K_{2}S{O}_4+2Cu{I}_2$
$2Cu{I}_2\to C{u}_2{I}_2+I_2$
$2N{a}_2{S}_2{O}_3+I_2\to 2NaI+N{a}_2{S}_4{O}_6$
$2moleCu\equiv 1mole{I}_1\equiv 2molehypo$
$20mL$ of solution reacts with $20mL$ of $0.0327N$ hypo
$\therefore 1000mL$ of solution will react with $1000mL$ of $0.0327N$ hypo
No, of moles of hypo used $=\frac{Mass}{Molecularmass\left(158\right)}$
$=\frac{E\times N\times V}{1000\times 158}$
where, $E=158,N=0.0327$ given, $V=1000mL$
$\therefore$ No. of moles of hypo used $=\frac{158\times 0.0327\times 1000}{1000\times 158}=0.0327$
No. of moles of $Cu=$ No. of moles of hypo $=0.0327mole$
Mass of copper in brass $=0.0327\times 63.5=2.07645$
% of copper in brass $=\frac{2.07645}{5}\times 100=41.529$%.