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Heat of Reaction

a CuSO4 rea...

Question

$(a)$ $C u S O_{4}$ reacts with $K I$ in acidic medium to liberate $I_{2}$ .
$2 C u S O_{4} + 4 K I ⟶ C u_{2} I_{2} + 2 K_{2} S O_{4} + I_{2}$
$(b)$ Mercuric perisdate, $H g_{5} (I O_{6})_{2}$ reacts with a mixture of $K I$ and $H C l$ according to the following the equation:
$H g_{5} (I O_{6})_{2} + 34 K I + 24 H C l ⟶ 5 K_{2} H g I_{4} + 8 I_{2} + 24 K C l + 12 H_{2} O$
$(c)$ The liberated iodine is titrated against $N a_{2} S_{2} O_{3}$ solution, one mL of which is equivalent to $0.0499$ g of $C u S O_{4} .5 H_{2} O$ .
What volume in mL of $N a_{2} S_{2} O_{3}$ solution will be required to react with $I_{2}$ liberated from $0.7245$ g of $H g_{5} (I O_{6})_{2}$ ? (Molar mass of $C u S O_{4} .5 H_{2} O = 249.5$ g/mol and of $H g_{5} (I O_{6})_{2} = 1448.5$ g/mol)

A

80

mL

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B

40

mL

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C

20

mL

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D

10

mL

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Solution

The correct option is B $40$ mL
Given that $1$ mol of $H g_{5} (I O_{6})_{2}$ or $1448.5$ g $= 8$ mol $I_{2}$

Moles of $H g_{5} (I O_{6})_{2} = \frac{0.7245}{1448.5}$

Moles of $I_{2}$ liberated $= \frac{8 \times 0.7245}{1448.5} = 4.0 \times 10^{- 3}$

The liberated $I_{2}$ is titrated against $N a_{2} S_{2} O_{3}$ .
$2 e + I_{2} ⟶ 2 I^{-}$
$2 (S^{2 +})_{2} ⟶ (S^{5 / 2})_{4} + 2 e$

Meq. of $N a_{2} S_{2} O_{3} =$ Meq. of $I_{2} = 2 \times 10^{3} \times$ mole of $I_{2}$
$= 4 \times 10^{- 3} \times 10^{3} \times 2 = 8$
Also, meq. of $N a_{2} S_{2} O_{3}$ in one mL $=$ Meq. of $C u S O_{4}$ $= \frac{0.0499}{\frac{249}{1}} \times 1000 = 0.20$
Meq. of $N a_{2} S_{2} O_{3}$ used $=$ Meq. of $I_{2}$
$0.20 \times V = 8$ $⟹ V = 40$ mL

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Similar questions

2.5

g sample of copper is dissolved in excess of

H_{2} S O_{4}

100

0.02 M C u S O_{4} (a q)

10

0.02 M

C u S O_{4} (a q)

is mixed with excess of

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to show the following changes:

C u S O_{4} + 2 K I ⟶ K_{2} S O_{4} + C u I_{2}

2 C u I_{2} ⟶ C u_{2} I_{2} + I_{2}

The liberated iodine is titrated with hypo (

N a_{2} S_{2} O_{3}

) which requires

V

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hypo solution for its complete reduction.The amount of

I_{2}

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10

0.02 M

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Q. What volume of

0.40 M - N a_{2} S_{2} O_{3}

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liberated by adding the excess KI to 50 mL of

0.20 M - C u S O_{4}

0.5

K_{2} C r_{2} O_{7}

K M n O_{4}

is treated with excess of

K I

in acidic medium.

I_{2}

liberated requires

100

0.15 N

N a_{2} S_{2} O_{3}

solution for titration.

%

K_{2} C r_{2} O_{7}

%

K M n O_{4}

2.5

g sample of copper is dissolved in excess of

H_{2} S O_{4}

100

0.02 M C u S O_{4} (a q)

10

0.02 M

C u S O_{4} (a q)

is mixed with excess of

K I

to show the following changes:

C u S O_{4} + 2 K I ⟶ K_{2} S O_{4} + C u I_{2}

2 C u I_{2} ⟶ C u_{2} I_{2} + I_{2}

The liberated iodine is titrated with hypo (

N a_{2} S_{2} O_{3}

) which requires

V

0.1 M

hypo solution for its complete reduction.

V

) of hypo required is:

2.5

g sample of copper is dissolved in excess of

H_{2} S O_{4}

100

0.02 M C u S O_{4} (a q)

10

0.02 M

C u S O_{4} (a q)

is mixed with excess of

K I

to show the following changes:

C u S O_{4} + 2 K I ⟶ K_{2} S O_{4} + C u I_{2}

2 C u I_{2} ⟶ C u_{2} I_{2} + I_{2}

The liberated iodine is titrated with hypo (

N a_{2} S_{2} O_{3}

) which requires

V

0.1 M

hypo solution for its complete reduction.

Percentage of purity of sample is:

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