One gram of commercial AgN O 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 is titrated with M/10 − KI

Question

One gram of commercial AgN O 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 is titrated with M/10 − KI O 3 ​ solution in the presence of 6 M - HCl till all iodide ions are converted into ICl . It requires 40 ml of M/10 − KI O 3 ​ solution. A 20 ml of the same stock solution of KI requires 24 ml of M/10 − KI O 3 ​ under similar conditions. The percentage of AgN O 3 ​ in the sample is (Ag = 108) Reaction: KI O 3 ​ + 2KI + 6HCl → 3ICl + KCl + 3 H 2 ​ O

TestHub · Accepted Answer

Blank: 20 ml stock \equiv 24 × 0.1 = 2.4 mmol I O 3 − ​ = 4.8 mmol I − per 20 ml, so 50 ml stock = 12 mmol I − . Leftover I − \equiv 40 × 0.1 = 4 mmol I O 3 − ​ = 8 mmol I − . I − removed = 12-8 = 4 mmol = mmol AgN O 3 ​ = 4 × 170 = 680 mg = 0.68 g. Purity = 68%. Ans 0068.

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