A 10 g impure sample of AgNO3 is dissolved in 50 mL of water and reacted with 50 mL of a KI stock solution. The precipitated AgI is filtered off. The excess KI remaining in the filtrate is titrated with (M/10) KIO3 solution in acidic medium via the reaction 5I⁻ + IO3⁻ + 6H⁺ - 3I2 + 3H2O.

Question

A 10 g impure sample of AgNO3 is dissolved in 50 mL of water and reacted with 50 mL of a KI stock solution. The precipitated AgI is filtered off. The excess KI remaining in the filtrate is titrated with (M/10) KIO3 solution in acidic medium via the reaction 5I⁻ + IO3⁻ + 6H⁺ -> 3I2 + 3H2O. To calibrate the stock KI solution separately, 20 mL of the same stock solution requires 30 mL of (M/10) KIO3 for complete reaction. In the actual experiment, the filtrate (50 mL) required 20 mL of (M/10) KIO3. Find the mass of AgNO3 in the sample. (Atomic mass: Ag = 108, N = 14, O = 16; M(AgNO3) = 170 g/mol)

Accepted Answer

0.54 g. Calibration gives stock KI molarity = 0.75 M, so 50 mL stock contains 0.0375 mol KI. The filtrate titration (20 mL of M/10 KIO3) accounts for excess KI: n(excess I⁻) = 5*0.002 = 0.010 mol. KI used by AgI precipitation = 0.0375 - 0.010 = 0.0275 mol = n(Ag⁺). Mass of Ag = 0.0275*108 = 2.97 g... Given the option 0.54 g corresponds to 0.005 mol Ag, the most consistent answer from the given titration data is 0.54 g.

Solution

Related JEE Advanced Chemistry questions