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One gram of commercial AgNO 3 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 filterate is titrated with (M/10) KIO­ 3 solution in presence of 6M HCI till all I - ions are converted into ICI. It requires 50 ml. of (M/10) KIO 3 solution. 20 ml. of the same stock solution of KI requires 30 ml. of (M/10) KIO 3 under similar conditions. Calculate the percentage of AgNO 3 in the sample. (Reaction : KIO 3 + 2KI + 6HCI → 3ICI + 3KCI + 3H 2 O)

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 filterate is titrated with (M/10) KIO­3 solution in presence of 6M HCI 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 + 6HCI → 3ICI + 3KCI + 3H2O)

Grade:upto college level

2 Answers

Navjyot Kalra
askIITians Faculty 654 Points
9 years ago
Reaction involved titration is
KIO3 + 2KI + 6HCI → 3 ICI + 3KCI + 3H2O
1 mole 2 mole
20 ml. of stock KI solution ≡ 30 ml. of M/10 KIO3 solution
Molarity of KI solution = 30 * 1 *2/20 *10 = 3/10
Millimoles in 50 ml. of KI solution = 50 * 3/10 = 15
Millimoles of KI left unreacted with AgNO3 solution
= 2 * 50 * 1/10 = 10
∴millimoles of KI reacted with AgNO3 = 15 – 10 = 5
Millimoles of AgNO3 present in AgNO3 solution = 5
∴ Wt. of AgNO3 in the solution = 5 * 10-3 * 170 = 0.850 g
% AgNo3 in the sample = 0.850/1 * 100 = 85%
Navjyot Kalra
askIITians Faculty 654 Points
9 years ago
Reaction involved titration is
KIO3 + 2KI + 6HCI → 3 ICI + 3KCI + 3H2O
1 mole 2 mole
20 ml. of stock KI solution ≡ 30 ml. of M/10 KIO3 solution
Molarity of KI solution = 30 * 1 *2/20 *10 = 3/10
Millimoles in 50 ml. of KI solution = 50 * 3/10 = 15
Millimoles of KI left unreacted with AgNO3 solution
= 2 * 50 * 1/10 = 10
∴millimoles of KI reacted with AgNO3 = 15 – 10 = 5
Millimoles of AgNO3 present in AgNO3 solution = 5
∴ Wt. of AgNO3 in the solution = 5 * 10-3 * 170 = 0.850 g
% AgNo3 in the sample = 0.850/1 * 100 = 85%

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