Question

$25$ mL of household bleach solution was mixed with $30$ mL of $0.50$ M $KI$ and $10$ mL of $4$ N acetic acid. In the titration of the liberated iodine, $48$ mL of $0.25$ N $N{a}_2{S}_2{O}_3$ was used to reach the end point. The molarity of the household bleach solution is :

• A. $0.48$ M
• B. $0.96$ M
• C. $0.24$ M
• D. $0.024$ M

Answer 1

The correct option is C $0.24$ M

Household bleach $+2KI\to I_2+$ products

$I_2+2N{a}_2{S}_2{O}_3\to N{a}_2{S}_4{O}_6+2NaI$

Amount of $N{a}_2{S}_2{O}_3$ used = $VM=(48\times {10}^{-3}L)\left(0.25mol{L}^{-1}\right)=12\times {10}^{-3}mol$

Amount of $I_2$ generated = $\frac{1}{2}(12\times {10}^{-3}mol)=6\times {10}^{-3}mol$

Assuming 1 mol of houshold bleach produces 1 mol $I_2$, we will have

Amount of houshold bleach in 25 mL solution = $6\times {10}^{-3}mol$

Molarity houshold bleach = $\frac{n}{V}=\frac{6\times {10}^{-3}mol}{25\times {10}^{-3}L}=0.24M$