(i) A solution of a mixture of KCl and KOH was neutralised with $120$ mL of $0.12$ N HCl. Calculate the amount of KOH in the mixture.
(ii) After titration, the resultant solution was made acidic with $H N O_{3}$ . Then excess of $A g N O_{3}$ , solution was added to precipitate the AgCl which weighed $3.7$ g after drying. Calculate percentage of KOH in the original mixture. $(0.806 g, 48.7 %)$ .

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Solution

(i) $(K C l + K O H) ⟶ n u e t r d l i s e d K C l + H_{2} O$
by $H C l$
$120 m l$ , $0.12 N$
$H C l$ neutrdlises only $K O H$
Hence $K O H = (120 \times 0.12)$ millimoles
$\Rightarrow 14.4$ millimoles
$= 0.8064 g$
(ii) After titration with solution made acidic with ${H N O}_{3}$
${A g N O}_{3} + K C l ⟶ A g C l$
$3.7 g$
$3.7 g$ $A g C l = 25.6$ millimoles
$25.6$ millimoles has $14.4$ millimoles from $K O H$
Hence $11.2$ millimoles from $K C l$ .
$11.2$ millimoles of $K C l \Rightarrow 0.846 g$ of $K C l$
Hence % $K O H = \frac{0.8064}{0.8064 + 0.846} \times 100$
$\approx 48.7$ %

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