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Types of Redox Reactions

A mixed solut...

Question

A mixed solution of potassium hydroxide and sodium carbonate required $15$ mL of $\frac{N}{20} H C l$ solution when titrated with phenolphthalein as an indicator. But the same amount of the solution when titrated with methyl orange as an indicator required $25$ ml of the same acid. The amount of $K O H$ present in the solution is :

A

0.014

g

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B

0.14

g

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C

0.028

g

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D

1.4

g

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Solution

The correct option is A $0.014$ g
Titration of sodium carbonate with $H C l$ by using methyl orange as the indicator requires twice the volume as required by titration with phenolphthalein indicator.
Thus, out of $15$ ml, the volume required for titration with sodium carbonate is $25 - 15 = 10$ ml.
The volume required for titration with $K O H$ is $15 - 10 = 5$ ml.
The number of millimoles of $K O H$ present in the solution is $\frac{5 \times 1}{20} = 0.25$ millimoles.
The amount of $K O H$ present in the solution is $\frac{0.25 \times 56.1}{1000} = 0.014$ g.

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Similar questions

Q. A mixed solution of potassium hydroxide and sodium carbonate required

15

\frac{N}{20} H C l

solution when titrated with phenolphthalein as an indicator. But the same amount of the solution when titrated with methyl orange as indicator required

25

mL of the same acid. The amount of

K O H

present in the solution is :

Q. A solution contains a mixture of

K O H

N a_{2} C O_{3}

. This solution requires

15

\frac{N}{20} H C l

when titrated with phenolphthalein as an indicator. However, the same amount of the solution when titrated with methyl orange as an indicator requires

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ml of the same acid. If the summation of weights of

K O H

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\frac{13.4}{100 x}

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56.1

N a_{2} C O_{3}

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Q. A solution containing

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N a O H

300 m L

0.1 N H C l

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Q. A 20 ml solution of 0.1 N

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