Question

1 gram of commercial $AgN{O}_3$ 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 $(M/10){\left(KIO\right)}_3$ solution in presence of $6MHCl$ till all $I-$ ions are converted into $ICl$ . It requires $50mLof(M/10){\left(KIO\right)}_3$ solution. $20mL$ of the same stock solution of $KI$ requires $30mLof(M/10){\left(KIO\right)}_3$ under similar conditions. Find the purity of $AgN{O}_3$ sample.

Reaction:
${\text{KIO}}_3+2\text{KI}+6\text{HCl}\to 3\text{ICl}+3\text{KCl}+3{\text{H}}_2\text{O}$

Answer 1

Answer: 85

Standardisation of KI solution
$\left[{\text{I}}^{-1}\to {\text{I}}^{+1}\text{Cl}+2{\text{e}}^{-}\right]\times 2$
${\text{I}}^{+5}{\text{O}}_3^{-1}+4{\text{e}}^{-}\to {\text{I}}^{+1}\text{Cl}$
$2{\text{I}}^{-}+{\text{IO}}_3^{-}+6{\text{H}}^{+}+3{\text{Cl}}^{-}\to 3\text{ICl}+3{\text{H}}_2\text{O}$
${\text{KIO}}_3=30\times \frac{1}{\text{10}}=\text{3mM}=\text{12 Meqs}$
$\text{= 12 Meqs = of KI}\cong \text{6 mM of KI}$ i.e. ${\text{M}}_{\text{KI}}=\text{0.3 M}$
$\cong \left(\text{20 ml}\right)\times \left(\frac{\text{3}}{10}\text{M}\right)\text{KI}$
Estimation of KI used with ${\text{AgNO}}_3$ solution
${\text{KIO}}_3\text{required for excess KI}$
$=\text{50}\times \frac{1}{10}=5\text{mM}\cong \text{20 Meqs}$

$=\frac{\text{20 Megs}}{2}=\text{10 mM KI.}$
(n factor for KI = 2)
$\text{Total KI used }{\text{with AgNO}}_3$
$=\text{50}\times \frac{3}{10}=\text{15 mM}$
KI used for precipitation
15 -10 = 5 mM
${\text{AgNO}}_3=\text{5 mM}=\text{5}\times 10^{-3}\times \text{170g/mole}$
$=\text{0.85g}$
% purely = 85%