JEE Advanced · Chemistry · 9. Redox Reactions
Bleaching powder and bleach solution are produced on a large scale and used in several household products. The effectiveness of bleach solution is often measured by iodometry.
Question:
\(25 \mathrm{~mL}\) of household solution was mixed with \(30 \mathrm{~mL}\) of \(0.50\) M KI and \(10 \mathrm{~mL}\) of \(4 \mathrm{~N}\) acetic acid. In the titration of the liberated iodine, \(48 \mathrm{~mL}\) of \(0.25 \mathrm{~N~} \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) was used to reach the end point. The molarity of the household bleach solution is
- A \(0.48 \mathrm{M}\)
- B \(0.96 \mathrm{M}\)
- C \(0.24 \mathrm{M}\)
- D \(0.024 \mathrm{M}\)
Answer & Solution
Correct Answer
(C) \(0.24 \mathrm{M}\)
Step-by-step Solution
Detailed explanation
Bleach \(+2 \mathrm{KI} \longrightarrow \mathrm{I}_{2}+\) Products
\(\mathrm{I}_{2}+2 \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3} \longrightarrow \mathrm{Na}_{2} \mathrm{~S}_{4} \mathrm{O}_{6}+2 \mathrm{NaI}\)
Number of millimole of hypo \(=0.25 \times 48\)
\(=2 \times\) millimole of \(\mathrm{I}_{2}\)
\(\therefore\) Number of millimole of \(\mathrm{I}_{2}=\frac{0.25 \times 48}{2}=6\)
millimole of \(\mathrm{I}_{2}=\) millimole of bleach
Molarity of bleaching solution
\(=\frac{\text { Millimoles of bleach }}{\text { Vol.(in } \mathrm{mL} \text { ) of bleach }}=\frac{6}{25}=0.24\)
\(\mathrm{I}_{2}+2 \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3} \longrightarrow \mathrm{Na}_{2} \mathrm{~S}_{4} \mathrm{O}_{6}+2 \mathrm{NaI}\)
Number of millimole of hypo \(=0.25 \times 48\)
\(=2 \times\) millimole of \(\mathrm{I}_{2}\)
\(\therefore\) Number of millimole of \(\mathrm{I}_{2}=\frac{0.25 \times 48}{2}=6\)
millimole of \(\mathrm{I}_{2}=\) millimole of bleach
Molarity of bleaching solution
\(=\frac{\text { Millimoles of bleach }}{\text { Vol.(in } \mathrm{mL} \text { ) of bleach }}=\frac{6}{25}=0.24\)
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