A $1.0 ext{ g}$ sample of commercial $ ext{AgNO}_3$ is dissolved in $50 ext{ mL}$ of water. It is

Question

A $1.0	ext{ g}$ sample of commercial $	ext{AgNO}_3$ is dissolved in $50	ext{ mL}$ of water. It is treated with $50	ext{ mL}$ of a $	ext{KI}$ solution. The silver iodide thus precipitated is filtered off. The excess $	ext{KI}$ in the filtrate is titrated with a $0.1	ext{ M KIO}_3$ solution in the presence of $6	ext{ M HCl}$ till all $	ext{I}^-$ ions are converted into $	ext{ICl}$. It requires $50	ext{ mL}$ of the $0.1	ext{ M KIO}_3$ solution. $20	ext{ mL}$ of the same stock solution of $	ext{KI}$ requires $30	ext{ mL}$ of $0.1	ext{ M KIO}_3$ under similar conditions. Calculate the percentage purity of $	ext{AgNO}_3$ in the sample. (Report as nearest integer. Reaction: $	ext{KIO}_3 + 2	ext{KI} + 6	ext{HCl} ightarrow 3	ext{ICl} + 2	ext{KCl} + 3	ext{H}_2	ext{O}$)

Professor PC Thomas Classes · Answer

From the stoichiometry of the given reaction, 1 mole of $	ext{KIO}_3$ reacts with 2 moles of $	ext{KI}$. 
Standardization of KI: $20	ext{ mL KI} \equiv 30	ext{ mL of } 0.1	ext{ M KIO}_3$. 
Moles of $	ext{KIO}_3 = 3	ext{ mmol} \implies$ Moles of $	ext{KI} = 6	ext{ mmol}$. 
Molarity of $	ext{KI} = 6/20 = 0.3	ext{ M}$. 
Total $	ext{KI}$ added to $	ext{AgNO}_3 = 50	ext{ mL} 	imes 0.3	ext{ M} = 15	ext{ mmol}$. 
Back titration of excess KI: required $50	ext{ mL of } 0.1	ext{ M KIO}_3 = 5	ext{ mmol KIO}_3$. 
Excess $	ext{KI} = 5 	imes 2 = 10	ext{ mmol}$. 
$	ext{KI}$ reacted with $	ext{AgNO}_3 = 15 - 10 = 5	ext{ mmol}$. 
$	ext{AgNO}_3$ and $	ext{KI}$ react in a 1:1 molar ratio, so moles of $	ext{AgNO}_3 = 5	ext{ mmol}$. 
Mass of $	ext{AgNO}_3 = 5 	imes 10^{-3} 	imes 170 = 0.85	ext{ g}$. 
Percentage purity $= (0.85/1) 	imes 100 = 85\%$.