Question

To a 25mL ${H_2}{O_2}$ solution, excess of acidified solution of KI was added. The iodine liberated required 20mL of 0.3N $N{a_2}{S_2}{O_3}$ solution. The volume strength of ${H_2}{O_2}$ solution is:
(A) 1.344 g/L
(B) 3.244 g/L
(C) 5.4 g/L
(D) 4.08 g/L

Answer 1

Hint: Two moles of sodium thiosulphate can reduce one mole of iodine. First calculate the moles of iodine produced. Then find the strength of the hydrogen peroxide solution with the help of reaction between ${H_2}{O_2}$ and $KI$.

Complete step by step solution:
We will first calculate the moles of the iodine generated and then we will calculate the strength of the hydrogen peroxide solution.
- The reaction between iodine and sodium thiosulphate can be given as
\[2N{a_2}{S_2}{O_3} + {I_2} \to N{a_2}{S_4}{O_6} + 2NaI\]
From the above reaction, we can say that two moles of sodium thiosulphate react with one mole of iodine.
So, we can say that 1000mL of 2N $N{a_2}{S_2}{O_3}$ reacts with 1 mole of iodine. Thus, 20mL of 0.3N $N{a_2}{S_2}{O_3}$ will react with $\dfrac{{20 \times 0.3 \times 1}}{{1000 \times 2}} = 0.003moles$ of iodine (${I_2}$).
Thus, we found that 0.003 moles of iodine gas would have been generated.
- Now, the reaction between KI and hydrogen peroxide to give iodine gas in an acidic medium can be given below.
\[2KI + {H_2}S{O_4} + {H_2}{O_2} \to {K_2}S{O_4} + 2{H_2}O + {I_2}\]
From the above reaction, we can say that the number of moles of iodine produced will be equal to the number of moles of ${H_2}{O_2}$ consumed.
So, the number of moles of ${H_2}{O_2}$ present in 25mL of its solution will be 0.003moles.
Now, we know that weight of 0.003moles of ${H_2}{O_2}$= $Number$ $of$ $moles$ $ \times $ $Molecular$ $weight$
Weight of ${H_2}{O_2}$ = $0.003$ $ \times $ $34$ $= 0.102g$
Now, we know that the strength of the solution is the weight of solute present in 1L of the solution. We know that 0.102g of ${H_2}{O_2}$ is present in 25mL of the solution.
So, strength of the ${H_2}{O_2}$ solution (weight of ${H_2}{O_2}$ in 1L solution) = $\dfrac{{1000 \times 0.102}}{{25}} = 4.08g/mL$
Thus, we obtained that strength of the solution is 4.08g/mL

So, the correct answer is (D).

Note: Remember that the volume strength of any solute is its weight in gm per one liter of the solution. Note that ${H_2}{O_2}$ acts as an oxidizing agent in the reaction with $KI$ where $KI$ acts as a reducing agent.