Some Basic Concepts of Chemistry 2 Question 39
39. One gram of commercial $\mathrm{AgNO}{3}$ is dissolved in $50 \mathrm{~mL}$ of water. It is treated with $50 \mathrm{~mL}$ of a KI solution. The silver iodide thus precipitated is filtered off. Excess of $\mathrm{KI}$ in the filtrate is titrated with $(\mathrm{M} / 10) \mathrm{KIO}{3}$ solution in presence of $6 \mathrm{M} \mathrm{HCl}$ till all $\mathrm{I}^{-}$ions are converted into ICl. It requires $50 \mathrm{~mL}$ of $(\mathrm{M} / 10) \mathrm{KIO}{3}$ solution, $20 \mathrm{~mL}$ of the same stock solution of $\mathrm{KI}$ requires $30 \mathrm{~mL}$ 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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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 $20 \mathrm{~mL}$ original $\mathrm{KI}$ solution $=3$ millimol.
$\Rightarrow 7.5$ millimol $\mathrm{KIO}_{3}$ would be required for original $50 \mathrm{~mL} \mathrm{KI}$.
$\Rightarrow$ Original $50 \mathrm{~mL} \mathrm{KI}$ solution contain 15 millimol of KI.
After $\mathrm{AgNO}{3}$ treatment 5 millimol of $\mathrm{KIO}{3}$ is required, i.e. 10 millimol $\mathrm{KI}$ is remaining.
$\Rightarrow 5$ millimol $\mathrm{KI}$ 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 $\mathrm{AgNO}_{3}=85 %$
