Question

2.5 g sample of bronze was dissolved in sulphuric acid to form $CuS{O}_4$. To this solution excess KI was added to form $CuI$ and $I_3^{-}$. The $I_3^{-}$ formed required 31.5 ml of 1.0 M sodium thiosulphote solution for completing titration. Determine the mass percentage of Cu in the Bronze sample (Cu: 63.5 I: 126.9)

Answer 1

The unbalanced reactions are
$Cu+H_{2}S{O}_4\to CuS{O}_4+H_{2}C{u}^{2+}+I^{-}\to CuI+I_3^{-}{I}_3^{-}+S_2{O}_3^{2-}\to I^{-}+S_4{O}_6^{2-}$

Since Iodine is involved in more than one product we cannot use the concepts of equivalence directly.
Balacing these equations:
$C{u}^{2+}+I^{-}\to CuI+I_3^{-}C{u}^{2+}\to C{u}^{+}{I}^{-}\to I_3^{-}$
Step 1:
$C{u}^{2+}+e^{-}\to C{u}^{+}3I^{-}\to I_3^{-}+2e^{-}$
Step 2:
$2C{u}^{2+}+3I^{-}\to 2C{u}^{+}+I_3^{-}$
Adding the counter ions:
$2C{u}^{2+}+3I^{-}\to 2CuI+I_3^{-}$
Step 3:
There are two more $I^{-}$ on the RHS and we need to balance this on the LHS
$2C{u}^{2+}+5I^{-}\to 2CuI+I_3^{-}...\left(1\right)$
This is the balanced equation
$I_3^{-}+S_2{O}_3^{2-}\to I^{-}+S_4{O}_6^{2-}$
We break this up as
$I_3^{-}\to I^{-}{S}_2{O}_3^{2-}\to S_4{O}_6^{2-}$
Balancing steps are as follows
Step 1:
$I_3^{-}+2e^{-}\to 3I^{-}2S_2{O}_3^{2-}\to S_4{O}_6^{2-}+2e^{-}...\left(2\right)$
We can just add up these equations to cancel electrons
$I_3^{-}+2S_2{O}_3^{2-}\to 3I^{-}+S_4{O}_6^{2-}$
We had earlier balanced the other equation,shown below
$2C{u}^{2+}+5I^{-}\to 2CuI+I_3^{-}...\left(1\right)$
From equations (1)and (2),1 mole of $C{u}^{2+}$ needs 1 mole of $S_2{O}_3^{2-}$
Also, 1 mole of Cu gives 1 mole of $C{u}^{2+}$
Hence moles of $S_2{O}_3^{2-}$= moles of $Cu$
Moles of $S_2{O}_3^{2-}=31.5\times 1=31.5mmol$
∴ Moles of $Cu=31.5mmol$
Molar mass of Cu = 63.5, ∴ mass of $31.5mmol=31.5\times {10}^{-3}\times 63.5=2g$
∴ mass percent $=\frac{2}{2.5}\times 100=80$ %