Question

5 g sample of brass was dissolved in one-liter $dil.H_{2}S{O}_4$. 20 mL of this solution was mixed with $KI,$ liberating $I_2$​ , and $C{u}^{+}$, and the $I_2$​ required 20 mL of 0.0327 N hypo solution for complete titration. Calculate the percentage of Cu in the alloy.

Answer 1

When brass is extracted with concentrated $H_{2}S{O}_4$, it gives copper sulphate.

$Cu+2H_{2}S{O}_4\to CuS{O}_4+S{O}_2+2H_{2}O$

$2CuS{O}_4+4KI\to 2K_{2}S{O}_4+2Cu{I}_2$

$2Cu{I}_2\to C{u}_2{I}_2+I_2$

$2N{a}_2{S}_2{O}_3+I_2\to 2NaI+N{a}_2{S}_4{O}_6$

$\text{2 moles Cu ≡ 1 mole}{I}_2=2\text{mole hypo}$

$\text{20 mL of solution reacts with 20 mL of 0.0327 N hypo}$

$\therefore \text{1000mL of the solution will react with 1000 mL of 0.0327 N hypo}$

$\text{No. of moles of hypo used}=\frac{Mass}{Molecularmass\left(158\right)}$

$=\frac{E\times N\times V}{1000\times 158}$

$\text{where},E=158,N=0.327given,V=1000mL$

$\therefore \text{No. of moles of hypo used}=\frac{158\times 0.0327\times 1000}{1000\times 158}=0.0327$

$\text{No. of moles of Cu=No. of moles of hypo =0.0327 mole}$

$\text{Mass of copper in brass}=0.0327\times 63.5=2.07645$

$\text{\% of copper in brass}=\frac{2.07645}{5}\times 100=41.529$%