NaIO3 reacts with NaHSO3 according to equationIO-3 - 3HSO-3 - I- - 3H- - 3SO42-The weight of NaHSO3 required to react with 100 mL of solution containing 0-68 g of NaIO3 is

MEq. of NaHSO_3 = mEq. of NaIO_3 = N × V=0.68198× 6 × 100 (I^5+ + 6e^ → I^+) ∴W_NaHSO_3M_NaHSO_3 nbsp;× 2 × 1000 = 0.68198× 6 × 1000 W_NaHSO_3 = 0.68 × 6 × 1 ...

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Standard XII

Chemistry

Introduction to Oxidation and Reduction

NaIO3 reacts ...

Question

$N a I O_{3}$ reacts with $N a H S O_{3}$ according to equation
$I O_{3}^{-} + 3 H S O_{3}^{-} \to I^{-} + 3 H^{+} + 3 S O_{4}^{2 -}$
The weight of $N a H S O_{3}$ required to react with $100 m L$ of solution containing $0.68 g$ of $N a I O_{3}$ is

5.2 g

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0.2143 g

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2.3 g

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None of the above

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Solution

The correct option is D None of the above
mEq. of $N a H S O_{3} =$ mEq. of $N a I O_{3} = N \times V = \frac{0.68}{198} \times 6 \times 100 (I^{5 +} + 6 e^{-} \to I^{+})$
$∴ \frac{W_{N a H S O_{3}}}{M_{N a H S O_{3}}} \times 2 \times 1000 = \frac{0.68}{198} \times 6 \times 1000$
$W_{N a H S O_{3}} = \frac{0.68 \times 6 \times 1000 \times 104}{198 \times 2 \times 1000} = 1.072 g$
Hence (d) is the correct answer.

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Similar questions

N a I O_{3}

reacts with

N a H S O_{3}

according to equation

I O_{3}^{-} + 3 H S O_{3}^{-} \to I^{-} + 3 H^{+} + 3 S O_{4}^{2 -}

The weight of

N a H S O_{3}

required to react with

100 m L

of solution containing

0.68 g

N a I O_{3}

N a I O_{3}

reacts with

N a H S O_{3}

according to equation

I O_{3}^{-} + 3 H S O_{3}^{-} \to I^{-} + 3 H^{+} + 3 S O_{4}^{2 -}

The weight of

N a H S O_{3}

required to react with

100 m L

of solution containing

0.68 g

N a I O_{3}

N a I O_{3}

reacts with

N a H S O_{3}

according to following equation

I O_{3}^{-} + 3 H S O_{3}^{-} \to I^{-} + 3 H^{+} + 3 S O_{4}^{2 -}

The equivalent weight of oxidant (molar mass =

M

) in the reaction is:

Chile salt petre is a natural source of

N a N O_{3}

which also contains

N a I O_{3}

. The

N a I O_{3}

can be used as a source of iodine produced in the following reactions:

I O_{3}^{-} + 3 H S O_{3}^{-} ⟶ I^{-} + 3 H^{+} 3 S O_{4}^{2 -}

... (i)

5 I^{-} + I O_{3}^{-} + 6 H^{+} ⟶ 3 I_{2} (s) + 3 H_{2} O

... (ii)

1.00

L of the starting solution which contains

5.80

N a I O_{3}

per litre is treated with stoichiometric quantity of

N a H S O_{3}

. Further, a quantity of the starting solution is added to the reaction mixture to bring about the second reaction.

[I = 127, N a = 23, S = 32]

How many grams of

N a H S O_{3}

are required in step (i)?

Q. Chile salt petre, a source of

N a N O_{3}

, also contains

N a I O_{3}

. The

N a I O_{3}

is a source of

I_{2}

produced as shown in the following equations:

Step I:

I O_{3}^{⊝} + 3 H S O_{3}^{⊝} \to I^{⊝} + 3 H^{⊝} + 3 S O_{4}^{2 -}

Step II:

5 I^{⊝} + I O_{3}^{⊝} + 6 H^{⊝} \to 3 I_{2} (s) + 3 H_{2} O

One litre sample of chile salt petre solution containing

6.6

N a I O_{3}

is treated with

N a H S O_{3}

. Now an additional amount of same solution is added to the reaction mixture to bring about the second titration. Calculate the weight of

N a H S O_{3}

required in step I and what additional volume of chile salt petre must be added in step II to bring out complete conversion of

I^{⊝}

I_{2}

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Introduction to Oxidation and Reduction

Standard XII Chemistry

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NaIO3 reacts with NaHSO3 according to equation IO−3+3HSO−3→I−+3H++3SO2−4 The weight of NaHSO3 required to react with 100 mL of solution containing 0.68 g of NaIO3 is

The correct option is D None of the above mEq. of NaHSO3= mEq. of NaIO3=N×V=0.68198×6×100 (I5++6e−→I+) ∴WNaHSO3MNaHSO3×2×1000=0.68198×6×1000 WNaHSO3=0.68×6×1000×104198×2×1000=1.072 g Hence (d) is the correct answer.

$N a I O_{3}$ reacts with $N a H S O_{3}$ according to equation
$I O_{3}^{-} + 3 H S O_{3}^{-} \to I^{-} + 3 H^{+} + 3 S O_{4}^{2 -}$
The weight of $N a H S O_{3}$ required to react with $100 m L$ of solution containing $0.68 g$ of $N a I O_{3}$ is