The percentage of copper in a copper(II) salt can be determined by using a thiosulphate titration. 0.305 g of a copper(II) salt was dissolved in water and added to an excess of potassium iodide solution liberating iodine according to the following equation: 2 Cu 2+( aq )+4 I -( aq ) leftharpoons 2 CuI ( s )+ I 2( aq ) The iodine liberated required 24.5 cm 3 of a 0.100 mol dm -3 solution of sodium thiosulphate. 2 S 2 O 32-( aq )+ I 2( aq ) longrightarrow 2 I -( aq )+ S 4 O 62-( aq ) The percentage of copper (by mass) in the copper(II) salt is [Atomic mass of copper =63.5 ]

Question

The percentage of copper in a copper(II) salt can be determined by using a thiosulphate titration. 0.305 g of a copper(II) salt was dissolved in water and added to an excess of potassium iodide solution liberating iodine according to the following equation: 2 Cu 2+( aq )+4 I -( aq ) leftharpoons 2 CuI ( s )+ I 2( aq ) The iodine liberated required 24.5 cm 3 of a 0.100 mol dm -3 solution of sodium thiosulphate. 2 S 2 O 32-( aq )+ I 2( aq ) longrightarrow 2 I -( aq )+ S 4 O 62-( aq ) The percentage of copper (by mass) in the copper(II) salt is [Atomic mass of copper =63.5 ] (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

2 × mol of I2= Mol of hypo. So mol of I 2=(24.5 × 0.1 × 10-3/2) Moles of Cu 2+=2 × (24.5 × 0.1 × 10-3/2)

2× mol of I2​= Mol of hypo. So mol of I2​=224.5×0.1×10−3​ Moles of Cu2+=2×224.5×0.1×10−3​

$2 \times$ mol of $I_{2} =$ Mol of hypo.
So mol of $I_{2} = \frac{24.5 \times 0.1 \times 1 0 ^{- 3}}{2}$
Moles of $C u^{2 +} = 2 \times \frac{24.5 \times 0.1 \times 1 0 ^{- 3}}{2}$