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Normality

How many mill...

Question

How many millilitres of 0.02 M $K M n O_{4}$ solution would be required to exactly titrate 25.0 mL of 0.2 M $F e (N O_{3})_{2}$ solution?

A

50.00 m L

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B

70.00 m L

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C

8 0.00 m L

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D

None of these

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Solution

The correct option is A $50.00 m L$
n factor of $K M n O_{4}$ ,

Equivalents of $K M n O_{4} = 5 \times$ moles of $K M n O_{4}$
Normality $K M n O_{4} = 5 \times$ molarity of $K M n O_{4}$

For $F e (N O_{3})_{2},$ moles and equivalents are equal, as n factor is 1.

At the equivalence point,

Equivalents of $K M n O_{4}$ = Equivalents of $F e (N O_{3})_{2}$

or $V_{K M n O_{4}} \times$ Normality of $K M n O_{4}$
$= V_{F e (N O_{3})_{2}} \times$ Normality of $F e (N O_{3})_{2}$

For 0.02 M $K M n O_{4}$ solution.

Normality of $K M n O_{4} = (5) (0.02000) = 0.1$ N

and for 0.20 M $F e (N O_{3})_{2}$ solution
Normality of $F e (N O_{3})_{2} = 0.20$ N

putting the values,
$V_{K M n O_{4}} = (25.00 m L) (\frac{0.2000}{0.1000}) = 50.00 m L$

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Similar questions

Q. How many millilitres of 0.02 M

K M n O_{4}

solution would be required to exactly titrate 25.0 mL of 0.2 M

F e (N O_{3})_{2}

0.46

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25.0

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Q. 25 ml of the given

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How many millilitres of 0.1 M $H_{2} S O_{4}$ must be added to 50 mL of 0.1 M $N a O H$ to give a solution that has a concentration of 0.05 M in $H_{2} S O_{4}$ ?

The following titration method is given to determine the total content of the species with variable oxidation states. Answer the question given at the end of it.

25.0

mL of solution containing both

F e^{2 +}

F e^{3 +}

ions is titrated with

25.0

0.02 M

K M n O_{4}

H_{2} S O_{4}

). As a result, all of the

F e^{2 +}

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F e^{3 +}

25

mL of the original solution is treated with

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