MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Introduction to Oxidation and Reduction

A 20 ml solut...

Question

A 20 ml solution of 0.1 N $N a_{2} C O_{3}$ is titrated with 0.1N HCl with phenolphthalein as the indicator. Another 20 ml solution of 0.1 N $N a_{2} C O_{3}$ is titrated with 0.1 N HCl with methyl orange as the indicator. The volume of HCl required for both the titrations resepectively is:

A

20 mL and 20 mL

No worries! We‘ve got your back. Try BYJU‘S free classes today!

B

10 mL and 20 mL

No worries! We‘ve got your back. Try BYJU‘S free classes today!

C

10 mL and 10 mL

No worries! We‘ve got your back. Try BYJU‘S free classes today!

D

20 mL and 10 mL

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D 20 mL and 10 mL
Methyl orange indicator changes color at the half equivalence point i.e conversion of sodium carbonate to sodium bicarbonate.
However phenolphthalein changes color at the complete neutralisation point.
Volume of sodium carbonate = 20 ml
Normality of sodium carbonate = 0.1 N
Normality of HCl = 0.1 N
Volume of HCl $= \frac{N_{N a O H} \times V_{N a O H}}{N_{H C l}}$
For complete neutralisation:
Volume of HCl = $\frac{20 \times 0.1}{0.1} = 20 m L$
For the half equivalence point volume = 10 mL

Suggest Corrections

thumbs-up

0

Similar questions

Q. A 20 ml solution of 0.1 N

N a_{2} C O_{3}

is titrated with 0.1N HCl with phenolphthalein as the indicator. Another 20 ml solution of 0.1 N

N a_{2} C O_{3}

is titrated with 0.1 N HCl with methyl orange as the indicator. The volume of HCl required for both the titrations resepectively is:

20 mL

0.1 M

solution of compound

N a_{2} C O_{3} \cdot N a H C O_{3} \cdot 2 H_{2} O

is titrated against

0.05 M H C l

x mL

H C l

is used when phenolphthalein is used as an indicator and

y mL

H C l

is used when methyl orange is the indicator in two separate titrations. Hence

(y - x)

Q. In the mixture of

N a H C O_{3} + N a_{2} C O_{3}

H C l

x

mL with phenolphthalein indicator and then

y

mL with methyl orange indicator in same titration. Hence, volume of

H C l

for complete reaction of

N a_{2} C O_{3}

50 mL

solution having

N a O H

N a_{2} C O_{3}

0.5 M 10 mL H C l

for complete titration with phenolphthalein indicator and another

50 mL

solution require

0.5 M, 12 mL H C l

solution in methyl orange indicator then what is the ratio of number of mole of

N a O H

N a_{2} C O_{3}

?

25 m L

of a mixture of

N a O H

N a_{2} C O_{3}

when titrated with

N / 10 H C l

using phenolphthalein indicator required

25 m L H C l

. The same volume of mixture when titrated with

N / 10 H C l

using methyl orange indicator required

30 m L

H C l

. Calculate the amount of

N a_{2} C O_{3}

N a O H

in one litre of this mixture.

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Introduction

CHEMISTRY

Watch in App

explore_more_icon

Explore more

Introduction to Oxidation and Reduction

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon