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Uses of Hydrogen

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Question

A solution containing $C u^{2 +}$ and $C_{2} O_{4}^{2 -}$ ions is titrated with $20$ mL of $\frac{M}{4}$ $K M n O_{4}$ solution in acidic medium. The resulting solution is treated with excess of $K I$ after neutralization. The evolved $I_{2}$ is then absorbed is $25$ mL of $\frac{M}{10}$ hypo solution. Which of the following statements is/are correct?

The difference in the number of mmol of

C u^{2 +}

and

C_{2} O_{4}^{2 -}

ions in the solution is

10

mmol

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The difference in the number of mmol of

C u^{2 +}

and

C_{2} O_{4}^{2 -}

ions in the solution is

22.5

mmol

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The equivalent weight of

C u^{2 +}

ions in the titration with

K I

is equal to the atomic weight of

C u^{2 +}

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The equivalent weight of

K I

in the titration is

\frac{M}{2}

where

M

is the molecular weight of

K I

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Solution

The correct options are
A The difference in the number of mmol of $C u^{2 +}$ and $C_{2} O_{4}^{2 -}$ ions in the solution is $10$ mmol
C The equivalent weight of $C u^{2 +}$ ions in the titration with $K I$ is equal to the atomic weight of $C u^{2 +}$
$A .$ $C u^{2 +}$ does not react with $M n O_{4}^{⊝}$ ion.
Only $C_{2} O_{4}^{2 -}$ reacts with $M n O_{4}^{⊝}$ ion.
$M n O_{4}^{⊝} (n = 5) = C_{2} O_{4}^{2 -} (n = 2)$
$m E q \equiv m E q$
$20 \times \frac{M}{4} \times 5 \equiv m E q$
$∴ m E q$ of $C_{2} O_{4}^{2 -} = 25$
mmoles of $C_{2} O_{4}^{2 -} = \frac{25}{2} = 12.5$
$B .$ $C u^{2 +} = K I = I_{2} = S_{2} O_{3}^{2 -}$ (hypo)
$m E q (n = 1) \equiv m E q \equiv m E q \equiv m E q (n = 1)$
$2 C u^{2 +} + 2 I^{⊝} \to C u_{2} I_{2}$ $(e + C u^{2 +} \to C u^{1 +})$
$∴ n = \frac{2}{2} = 1$
$2 S_{2} O_{3}^{2 -} \to S_{4} O_{6}^{2 -} + 2 e^{-}$
mEq of $S_{2} O_{3}^{2 -} \equiv 25 \times \frac{M}{10} \times 1$
$= 2.5 = m E q$ of $C u^{2 +}$
mEq of $C u^{2 +} = 2.5$
mmoles of $C u^{2 +} = \frac{2.5}{1} = 2.5$
Difference in mmoles of $C_{2} O_{4}^{2 -}$ and $C u^{2 +} = 12.5 - 2.5 = 10$
Equivalent weight of $C u^{2 +} = \frac{A t o m i c w e i g h t o f C u^{2 +}}{1 (n f a c t o r = 1)}$
Equivalent weight of $K I = \frac{M}{n - f a c t o r} = \frac{M}{1} = M ⎛ ⎝ \begin{matrix} 2 I^{⊝} \to I_{2} + 2 e^{-} n = \frac{2}{2} = 1 \end{matrix} ⎞ ⎠$
Hence option $A$ & $C$ are correct.

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Similar questions

Q. A solution containing

C u^{2 +}

and

C_{2} O_{4}^{-}

ions which is titrated with

20 m l

M / 4 K M n O_{4}

solution in acidic medium. The resulting solution is treated with excess of

K I

after neutralization. The evolved

I_{2}

is then absorbs in

25 m l

N / 10

hypo solution, then the difference of the number millimoles of

C u^{2 +}

and

C_{2} O_{4}^{2 -}

ions in the solution is:

Q. A solution of

0.2 g

of a compound containing

C u^{2 +}

and

C_{2} O_{4}^{2 -}

ions on titration with

0.02 M

K M n O_{4}

in the presence of

H_{2} S O_{4}

consumes

22.6

mL of the oxidant.

The resultant solution is neutralised with

N a_{2} C O_{3}

, acidified with dilute acetic acid and treated with excess

K I

. The liberated iodine requires

11.3

mL of

0.05

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

solution for complete reduction. The mole ratio of

C u^{2 +}

and

C_{2} O_{4}^{2 -}

in the compound will be :

Q. A solution of 0.2 g of a compound containing

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and

C_{2} O_{4}^{2 -}

ions on titration with 0.02 M

K M n O_{4}

in the presence of

H_{2} S O_{4}

consumes 22.6 mL of the oxidant. The resultant solution is neutralized with

N a_{2} C O_{3}

, acidified with dilute acetic acid and treated with excess of

K I

. The liberated iodine requires 11.3 mL of 0.05 N

N a_{2} S O_{4}

solution for complete reduction . Find out the molar ratio of

C u^{2 +}

C_{2} O_{4}^{2 -}

in the compound.

Q. A solution of

0.2 g

of a compound containing

C u^{2 +}

and

C_{2} O_{4}^{2 -}

ions on titration with

0.02 M K M n O_{4}

in presence of

H_{2} S O_{4}

consumes

22.6 m L

of the oxidant. The resultant solution is neutralized with

N a_{2} C O_{3}

, acidified with dil. acetic acid and treated with excess

K I

.The liberated iodine requires

11.3 m L

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

solution for complete reduction. Find out the mole ratio of

C u^{2 +}

C_{2} O_{4}^{2 -}

in the compound is

a : b

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a + b

Q. A

0.276

g impure sample of copper ore is dissolved and

C u^{2 +}

is titrated with

K I

solution.

I_{2}

liberated required

40

mL of

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

solution for titration. The

%

of impurities in the ore is:

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A solution containing Cu2+ and C2O2−4 ions is titrated with 20 mL of M4 KMnO4 solution in acidic medium. The resulting solution is treated with excess of KI after neutralization. The evolved I2 is then absorbed is 25 mL of M10 hypo solution. Which of the following statements is/are correct?