MyQuestionIcon

1

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

Common Ion Effect

25 mL solutio...

Question

$25$ mL solution containing a mixture of $N a_{2} C O_{3}$ and $N a O H$ required $19.5$ mL of $0.995 N$ $H C l$ for the phenolpthalein end-point and in a separate titration, $25$ mL of the same mixture required $25.9$ mL of same $H C l$ for methyl orange end-point. The concentration (in g/L) of $N a_{2} C O_{3}$ in the mixture is $9 X$ . Find the value of $X$ .

Open in App

Solution

$25.9 - 19.5 = 6.4 m L$ of $0.995 N H C l$ corresponds to the half neutralization of $N a_{2} C O_{3}$ .
The number of moles of $N a_{2} C O_{3}$ is $0.995 N \times 6.4 \times 10^{- 3} = 6.368 \times 10^{- 3} m o l e s$
The concentration of $N a_{2} C O_{3}$ is $\frac{6.368 \times 10^{- 3} m o l e s \times 106 g / m o l}{25 \times 10^{- 3} L} = 27 g / L$ .
value of $X$ = $3$
$19.5 - 6.4 = 13.1 m L$ of $0.995 N H C l$ corresponds to the neutralization of $N a O H$ .
The number of moles of $N a O H$ is $0.995 N \times 13.1 \times 10^{- 3} = 13.03 \times 10^{- 3} m o l e s$
Thus, the concentration of $N a O H$ is $\frac{13.03 \times 10^{- 2} m o l e s \times 40 g / m o l}{25 \times 10^{- 3} L} = 20.86 g / L$

Suggest Corrections

thumbs-up

1

Similar questions

200

N a O H

N a_{2} C O_{3}

was first titrated with phenolpthalein and

17.5

H C l

with normality 0.10

N

was required for end point. After this, methyl orange was added and

2.5

H C l

was again required for next end point. Find out the amounts of

N a O H

N a_{2} C O_{3}

in the mixture.

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}

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. Given two mixtures: (A)

N a O H

N a_{2} C O_{3}

(B)

N a H C O_{3}

N a_{2} C O_{3}

.

100

mL of mixture (A) required

a

b

1 M

H C l

in separate titration using phenolphthalein and methyl orange indicators respectively while

100

mL of mixture (B) required

x

y

H C l

solution is separate titration using the same indicators.

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Degree of Dissociation

CHEMISTRY

Watch in App

explore_more_icon

Explore more

Common Ion Effect

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon