Chemistry - Redox Reactions Question with Solution | TestHub
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 is titrated with M/10-KIO₃ solution in the presence of 6 M - HCl till all iodide ions are converted into ICl. It requires 50 ml of M/10-KIO₃ solution. A 20 ml of the same stock solution of KI requires 30 ml of M/10-KIO₃ under similar conditions. The percentage of AgNO₃ in the sample is (Ag = 108) Reaction: KIO₃ + 2KI + 6HCl 3ICl + KCl + 3H₂O
Answer:
Solution:
Blank: 20 ml KI stock \equiv 30×0.1 = 3 mmol IO₃⁻; with IO₃⁻ : I⁻ = 1:2, stock holds 6 mmol I⁻ per 20 ml, so 50 ml holds 15 mmol. After AgI precipitation, leftover I⁻ \equiv 50×0.1 = 5 mmol IO₃⁻ = 10 mmol I⁻. I⁻ removed by Ag⁺ = 15-10 = 5 mmol = mmol AgNO₃ = 5×170 = 850 mg = 0.85 g. Purity = 85%. Ans 0085.
