Question

$0.1gKI{O}_3$ and excess $KI$ when treated with $HCl$, the iodine is liberated. The liberated iodine required $45mL$ sodium thiosulphate for titration. The molarity of sodium thiosulphate will be equivalent to:

• A. $0.252M$
• B. $0.126M$
• C. $0.0313M$
• D. $0.0623M$

Answer 1

The correct option is D $0.0623M$

Given,

$0.1gKI{O}_3,$ molar mass of $KI{O}_3=214$

$\therefore$ Molar of $KI{O}_3=\frac{Givenmass}{Molarmass}=\frac{0.1}{214}$

$=0.00047moles$

Now,

$KI{O}_3+5KI+3H_{2}S{O}_4\rightleftharpoons 3K_{2}S{O}_4+3H_{2}O+3I_2\uparrow$

$I_2$ is liberated from $KI{O}_3$,

$\Rightarrow 0.00047÷2=0.000235$

Now, moles of KI reacting $=0.00047\times 5$

$=0.000235$

$\therefore$ moles of $I_2$ Produced from $KI=0.000235÷2$

$=0.0001175$

$\therefore$ Total moles of $I_2=0.000235+0.0001175$

$=0.000141$ equivalent of $I_2$

Equivalent of Thiosulphate $=2\times 0.000141$

$=0.000282$

$\therefore$ Morality $=0.000282\times \frac{1000}{45}$

$=0.00626M$