Question

How many mL of 0.1411 M thiosulfate titrant do you expect to need to titrate a fully dissolved and treated 0.3045 g sample of brass containing 86.87% copper (M 63.546) that has been reacted with excess KI?  Report your result to 2 decimal places, since that is how you should be reading a buret.

The reactions are:

I₂ + 2S₂O₃^2_- --> products we don't care about

2Cu²⁺ + excessI^- --> I₂

Answer 1

The reaction between brass ( Cu⁺² is the main reactant) and excess KI is shown below:

2 Cu⁺² + 2 I^- = 2 Cu⁺ + I₂

From the balanced reaction stoichiometry, we get:

Moles of I₂ / Moles of Cu⁺² = 1/2

So, moles of I₂ = (1/2) * Moles of Cu⁺²

The mass of Cu⁺² present in 0.3045 g of brass sample = 0.3045 g * (86.87/100) = 0.2645 g

Moles of Cu⁺² present = mass of Cu⁺² / atomic mass of Cu⁺² = 0.2645 g / (63.546 g/ mole) = 0.00416 mole

From the stoichiometry, we get:

Moles of I₂ produced = (1/2) * moles of Cu+2 reacted = (1/2) * 0.00416 mole = 0.00208 mole

So, 0.00208 moles of I₂ is titrated by thiosulphate solution.

The balanced equation between I₂ and S₂O₃^-2 is:

I₂ + 2S₂O₃^-2 = S₄O₆^-2 + 2 I^-

So, moles of S₂O₃^-2 / moles of I₂ = 2/1

Or, moles of S₂O₃^-2 = 2 * moles of I₂

Putting the value of moles of I2 = 0.00208 mole in the above equation, we get:

moles of S₂O₃^-2 = 2 * 0.00208 mole = 0.00416 mole

So, Volume of the thiosulfate solution needed for titration = Moles of thiosulphate / molarity of thiosulphate = ( 0.00416 mole/ 0.1411 mole/Liter) = 0.0295 L = 0.0295 L * (1000 ml/ 1 L) = 29.5 mL

The required volume of thiosulphate is : 29.5 mL.