Question

One gram of commercial AgNO₃ is dissolved in 50 mL of water. It is treated with 50 mL of a KI solution. The silver iodide thus precipitated is filtered off. Excess of KI in the filtrate is titrated with (M/10) KIO₃ solution in presence of 6 M HCl till all I^- ions are converted into ICI. It requires 50 mL of (M/10) KIO₃ solution. 20 mL of the same stock solution of KI requires 30 ml of (M/10) KIO₃ under similar conditions. Calculate the percentage of AgNO₃ in the sample.

Reaction

KIO₃ + 2KI + 6HCl → 3ICl + 3KCl + 3H₂O

Answer 1

Number of millimoles of KIO₃ in 30 mL of solution = molarity x volume in mL

= 1/10 x 30 = 3

Given equation :

KIO₃ + 2KI + 6HCl → 3ICl + 3KCl + 3H₂O

According to the equation of the reaction given, 1 mole of KIO₃ is equivalent to 2 moles of KI

:. No. of millimoles of KI in 20 mL of stock solution

= 2 x 3 = 6

:. No. of millimoles of KI in 50 mL of the same solution

= 6 x 50/20 = 15

No. of millimoles of KIO₃ in 50 mL of solution

= 1/10 x 50 = 5

:. No. of millimoles of KI used with 50 mL of KIO₃ solution = 2 x 5 = 10

:. No. of millimoles of Kl used with AgNO₃ 

= 15 - 10 = 5

AgNO₃ + KI → AgI + KNO₃

1 mole of AgNO₃ reacts with 1 mole of KI. Therefore, number of millimoles of AgNO₃ is equal to 5.