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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Chemistry
Chapter
Redox Reactions

NTA Abhyas 2020 · Chemistry · Redox Reactions

25mL of household bleach solution was mixed with 30mL of 0.50MKI...

25mL25mL of household bleach solution was mixed with 30mL30mL of 0.50MKI0.50MKI and 10mL10mL of 4N4N acetic acid. In the titration of the liberated iodine, 48mL48mL of 0.25NNa2S2O30.25NN{a}_{2}{S}_{2}{O}_{3} was used to reach the end point. The molarity of the household bleach solution is

Options

  1. A0.24 M
  2. B0.48
  3. C0.024 M
  4. D0.96 M

Correct answer

A. 0.24 M

Step-by-step solution

\mathop{{H}_{2}{O}_{2}}_{\mathop{\text{Bleach}_{\text{25 ml}}} + \mathop{\text{2}{\textI^{-}}_{\mathop{0. \text{5M}_{30\text{ml}}} + \mathop{\text{2}{\textH^{+}}_{\mathop{( \text{from C}{\text{H}_{3}\text{COOH} )}_{\text{4 N} ( 10\text{ml} )}} \to {I}_{2} + \text{2}{\text{H}_{2}O...(i)

{I}_{2} + \mathop{\text{2N}{\text{a}_{2}{S}_{2}{O}_{3}}_{0. \text{25 N} ( \text{48 ml} )} \to \text{N}{\text{a}_{2}{S}_{4}{O}_{6} + \text{2NaI}...(ii)

m. mol of I2=12{I}_{2} = \dfrac{1}{2} (m. Moles of \text{N}{\text{a}_{2}{S}_{2}{O}_{3} )

=12×(0.25×48)=6 m.mol= \dfrac{1}{2} \times ( 0. \text{25} \times \text{48} ) = \text{6 m}. \text{mol}

Using equation (i)

1 m. Mol of I2≡1{I}_{2} ≡ 1 m. Mol of H2O2{H}_{2}{O}_{2}

m.mol of H2O2=6{H}_{2}{O}_{2} = 6 m.mol

Molarity of H2O2=625=0.24M{H}_{2}{O}_{2} = \dfrac{6}{25} = 0.24M
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