Some Basic Concepts of Chemistry - Result Question 36
39. One gram of commercial $\mathrm{AgNO}_3$ 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 filtrate is titrated with $(\mathrm{M} / 10) \mathrm{KIO}_3$ solution in presence of 6 M HCl till all $\mathrm{I}^{-}$ions are converted into ICl . It requires 50 mL of $(\mathrm{M} / 10) \mathrm{KIO}_3$ solution, 20 mL of the same stock solution of KI requires 30 mL of $(\mathrm{M} / 10) \mathrm{KIO}_3$ under similar conditions. Calculate the percentage of $\mathrm{AgNO}_3$ in the sample.
Reaction $\mathrm{KIO}_3+2 \mathrm{KI}+6 \mathrm{HCl} \longrightarrow 3 \mathrm{ICl}+3 \mathrm{KCl}+3 \mathrm{H}_2 \mathrm{O}$
(1992, 4M)*
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Answer:
Correct Answer: 39. (85%)
Solution:
- The reaction is
$\mathrm{KIO}_3+2 \mathrm{KI}+6 \mathrm{HCl} \longrightarrow 3 \mathrm{ICl}+3 \mathrm{KCl}+3 \mathrm{H}_2 \mathrm{O}$
$\mathrm{KIO}_3$ required for 20 mL original KI solution $=3$ millimol.
$\Rightarrow 7.5$ millimol $\mathrm{KIO}_3$ would be required for original 50 mL KI .
$\Rightarrow$ Original 50 mL KI solution contain 15 millimol of KI.
After $\mathrm{AgNO}_3$ treatment 5 millimol of $\mathrm{KIO}_3$ is required, i.e. 10 millimol KI is remaining.
$\Rightarrow 5$ millimol KI reacted with 5 millimol of $\mathrm{AgNO}_3$.
$\Rightarrow$ Mass of $\mathrm{AgNO}_3=\frac{5}{1000} \times 170=0.85 \mathrm{~g}$
$\Rightarrow$ Mass percentage of $\mathrm{AgNO}_3=85 %$
BETA
