MyQuestionIcon

1

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

Types of Redox Reactions

A mixed solut...

Question

A mixed solution of potassium hydroxide and sodium carbonate required $15$ mL of an $\frac{N}{20} H C l$ 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 :

A

0.014

g

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

B

0.14

g

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

C

0.028

g

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

D

0.25

g

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

Open in App

Solution

The correct option is A $0.014$ g
$K O H (x m m o l) + N a_{2} C O_{3} (y m m o l)$
i. $(x \times 1) + (y \times 2) \times \frac{1}{2} = \frac{1}{20} \times 15$

ii. $(x \times 1) + (y \times 2) = \frac{1}{20} \times 25 \Rightarrow y = 0.5$ and $x = 0.25$

$\Rightarrow K O H \equiv x$ mmoles $= 0.25 \times 10^{- 3} \times 56 g = 0.014 g$

Suggest Corrections

thumbs-up

0

Similar questions

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

25

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

K O H

N a_{2} C O_{3}

in the solution is

x

then find the value of

\frac{13.4}{100 x}

(Given that the molecular mass of

K O H

56.1

N a_{2} C O_{3}

106

Q. A solution containing

N a_{2} C O_{3}

N a O H

300 m L

0.1 N H C l

using phenolphthalein as an indicator. Methyl orange is then added to the above-titrated solution when a further

25 m L

0.2 N H C l

is required. The amount of

N a O H

present in the original solution is:

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:

500 m L

of a solution of

C a C O_{3}

N a O H

was first titrated with

\frac{N}{10} H C l

using phenolphthalein indicator.

18.5 m L

H C l

was required for the end point. After that methyl orange indicator was used and

2.5 m L

H C l

was required for the next end point. Find out the mass of

C a C O_{3}

in the mixture.

Q. A solution containing

N a_{2} C O_{3}

N a O H

300 m l

0.1 N H C l

using phenolphthalein as indicator. Methyl orange is then added to the above solution when

25 m l

0.2 N H C l

is required. The amount of

N a O H

present in the solution is:

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Types of Redox Reactions

CHEMISTRY

Watch in App

explore_more_icon

Explore more

Types of Redox Reactions

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon