Question

1 gram of commercial $AgNO_{3}$ is made into solution of 50 mL and $KI$ is added in excess. The silver iodide thus precipitated is filtered off. Excess of $KI$ in the filtrate is titrated with $\left(\right.M/10\left.\right)\left(KIO\right)_{3}$ solution in presence of $6MHCl$ till all $I-$ ions are converted into $ICl$ . It requires $50mLof\left(\right.M/10\left.\right)\left(KIO\right)_{3}$ solution. $20mL$ of the same stock solution of $KI$ requires $30mLof\left(\right.M/10\left.\right)\left(KIO\right)_{3}$ under similar conditions. Find the purity of $AgNO_{3}$ sample.

Reaction:
$\text{KIO}_{3} + 2 \text{KI} + 6 \text{HCl} \rightarrow 3 \text{ICl} + 3 \text{KCl} + 3 \text{H}_{2} \text{O}$

Answer 1

Standardisation of KI solution
$\left[\text{I}^{-1} \rightarrow \text{I}^{+ 1} \text{Cl} + 2 \text{e}^{-}\right] \times 2$
$\text{I}^{+ 5} \text{O}_{3}^{- 1} + 4 \text{e}^{-} \rightarrow \text{I}^{+ 1} \text{Cl}$
$2 \text{I}^{-} + \text{IO}_{3}^{-} + 6 \text{H}^{+} + 3 \text{Cl}^{-} \rightarrow 3 \text{ICl} + 3 \text{H}_{2} \text{O}$
$\text{KIO}_{3} = 3 0 \times \frac{1}{\text{10}} = \text{3mM} = \text{12 Meqs}$
$ \text{= 12 Meqs = of KI} \cong \text{6 mM of KI}$ i.e. $\text{M}_{\text{KI}} = \text{0.3 M}$
$\cong \left(\text{20 ml}\right) \times \left(\frac{\text{3}}{1 0} \text{M}\right) \text{KI}$
Estimation of KI used with $\text{AgNO}_{3}$ solution
$\text{KIO}_{3} \text{required for excess KI}$
$= \text{50} \times \frac{1}{1 0} = 5 \text{mM} \cong \text{20 Meqs}$

$= \frac{\text{20 Megs}}{2} = \text{10 mM KI.}$
(n factor for KI = 2)
$\text{Total KI used } \text{with AgNO}_{3}$
$= \text{50} \times \frac{3}{1 0} = \text{15 mM}$
KI used for precipitation
15 -10 = 5 mM
$\text{AgNO}_{3} = \text{5 mM} = \text{5} \times 1 0^{- 3} \times \text{170g/mole}$
$= \text{0.85g}$
% purely = 85%