Ten grams of an impure AgNO 3 sample is dissolved in 50 mL of water. It is treated with 50 ml of KI solution. The silver iodide thus precipitated is filtered off. Excess of KI in the filterate is ti

Question

Ten grams of an impure AgNO 3 ​ sample is dissolved in 50 mL of water. It is treated with 50 ml of KI solution. The silver iodide thus precipitated is filtered off. Excess of KI in the filterate is titrated with 50 ml of (M/10) KIO 3 ​ solution in presence of acidic medium ( I − + IO 3 − ​ → I 2 ​ ) .20 ml of the same stock solution of KI requires 30 mL of ( M /10 ) KIO 3 ​ under similar condition in different experiment. Calculate the percentage of AgNO 3 ​ in the sample. (Take molar mass of Ag = 108, N = 14, O = 16)

TestHub · Accepted Answer

Question Explanation: Calculate the purity of an $A g N O_{3}$ sample using back-titration of excess KI with $K I O_{3}$ .

Concept: Equivalents of titrant equal equivalents of analyte.

Solution: First, find the molarity (M) of the KI solution from the second experiment:

mEq of $K I = m Eq$ of $K I O_{3}$ (n-factor of $I^{-} \to I_{2}$ is 1, $I O_{3}^{-} \to I_{2}$ is 5)

$20 \times M \times 1 = 30 \times 0.1 \times 5 \Rightarrow M = 0.75 m o l / L$

Now, calculate millimoles of $A g^{+} (x)$ by finding the excess KI in the first titration:

mEq of excess $K I = m Eq$ of $K I O_{3}$

$(50 \times 0.75 - x) \times 1 = 50 \times 0.1 \times 5 \Rightarrow 37.5 - x = 25 \Rightarrow x = 12.5 mm o l$

Calculate the mass and percentage of $A g N O_{3}$ (Molar mass $= 170$ ):

Mass $= 12.5 \times 1 0^{- 3} \times 170 = 2.125 g$

$P u r i t y = \frac{2.125}{10} \times 100 = 21.25%$

Final Answer: 21.25%

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JEE Advanced 2028 (Class 11)