MyQuestionIcon

1

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

Equivalent Mass

A solution co...

Question

A solution contains a mixture of $N a_{2} C O_{3}$ and $N a O H .$ Using phenolphthalein as indicator $35 m L$ of mixture required $20 m L$ of $0.80 N H C l$ for the end point. With methyl orange, $35 m L$ of solution required $37.5 m L$ of the same $H C l$ for the end point. What is the weight of $N a_{2} C O_{3}$ and $N a O H$ respectively in the mixture?

A

1.484 g, 0.08 g

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

B

1.5 g, 0.5 g

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

C

0.08 g, 1.484 g

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

D

0.5 g, 1.5 g

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

Open in App

Solution

The correct option is A $1.484 g, 0.08 g$
Suppose milliequivalents of $N a_{2} C O_{3}$ and $N a O H$ are $x$ and $y$ respectively.

When phenolphthalein is used as an indicator, only $50 %$ of $N a_{2} C O_{3}$ will be neutralised.
$\frac{1}{2} Milliequivalents of N a_{2} C O_{3} + Milliequivalents of N a O H = Milliequivalents of H C l$
$∴ \frac{x}{2} + y = 20 \times 0.80$
$\frac{x}{2} + y = 16 . . . . . (1)$

When methyl orange is used as an indicator, $100 %$ of $N a_{2} C O_{3}$ will be neutralised.
$Milliequivalents of N a_{2} C O_{3} + Milliequivalents of N a O H = Milliequivalents of H C l$
$∴ x + y = 37.5 \times 0.80$
$x + y = 30 . . . . . (2)$

From equation $(1)$ and $(2)$ ,
$x = 28, y = 2$

$∴ Milliequivalents of N a_{2} C O_{3} = 28.0$
$\frac{W_{N a_{2} C O_{3}}}{M_{N a_{2} C O_{3}}} \times 2 \times 1000 = 28.0$
$∴ W_{N a_{2} C O_{3}} = \frac{28.0 \times 106}{2 \times 1000} = 1.484 g$

$∴ Milliequivalents of N a O H = 2.0$
$\frac{W_{N a O H}}{M_{N a O H}} \times 1 \times 1000 = 2.0$
$∴ W_{N a O H} = \frac{2.0 \times 40}{1 \times 1000} = 0.08 g$

Suggest Corrections

thumbs-up

1

Similar questions

Q. A solution contains a mixture of

N a_{2} C O_{3}

N a O H .

Using phenolphthalein as indicator

35 m L

of mixture required

20 m L

0.80 N H C l

for the end point. With methyl orange,

35 m L

of solution required

37.5 m L

H C l

for the end point. What is the weight of

N a_{2} C O_{3}

N a O H

respectively in the mixture?

Q. 200 mL of a solution mixture

N a O H

N a_{2} C O_{3}

was first titrated with phenolphthalein and N/10 HCl 17.5 mL of HCl was required for the end point. After this methyl orange was added and 2.5 mL of same HCl was again required for next end point. Find out the mass of NaOH and

N a_{2} C O_{3}

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

20 m l

H C l

using phenolphthalein as indicator and

35 m l

more required if methyl orange is used as indicator. Then molar ratio of

N a H C O_{3}

N a_{2} C O_{3}

Q. A solution contains

{N a}_{2} C O_{3}

N a O H

. Using phenolphthalein as an indicator,

25

mL of a mixture requires

19.5

0.005 N H C l

for the end point. If methyl orange is the indicator, then

25

mL of the solution requires

25.9

H C l

for end point. The concentration of

N a O H

substance in g per litre is (write it as

0.23 = 2

(only single digit)) :

Q. A mixture of NaOH and

N a_{2} C O_{3}

required 25mL of 0.21 m HCl using phenolphthalein as the indicator. However, the same amount of the mixture required 30 mL of 0.1 M HCl when methyl orange was used as the indicator. The molar ratio of NaOH and

N a_{2} C O_{3}

in the mixture was :

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Reactions in Solutions

CHEMISTRY

Watch in App

explore_more_icon

Explore more

Equivalent Mass

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon