MyQuestionIcon

1

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

Hydrochloric Acid

A solution co...

Question

A solution contains ${N a}_{2} C O_{3}$ and $N a O H$ . Using phenolphthalein as an indicator, $25$ mL of a mixture requires $19.5$ mL of $0.005 N H C l$ for the end point. If methyl orange is the indicator, then $25$ mL of the solution requires $25.9$ of the same $H C l$ 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)) :

Open in App

Solution

Meq. of $H C l$ for $35$ mL mixture using phenolphthalein indicator = $19.5 \times 0.005 = 0.0975$
$∴$ Meq. of $N a O H +$ $\frac{1}{2}$ Meq. of ${N a}_{2} C O_{3} = 0.0975$ ...(i)
Now, meq. of $H C l$ used for $25$ mL mixture using methyl orange as indicator $= 25.9 \times 0.005 = 0.1295 . . . (i i)$
$∴$ By eqs. $(i)$ and $(i i)$ , we get meq. of $N a O H$ $= 0.0655$
$\frac{w}{40} \times 1000 = 0.0655$
$∴$ Mass of $N a O H =$ $2.62 \times 10^{- 3} g / 25$
Strength of NaOH = $0.1048 g / l i t r e$
and answer is $1$ .
Also,
Meq. of ${N a}_{2} C O_{3} = 0.064$
$\frac{w}{106 / 2} \times 1000 = 0.064$
Mass of ${N a}_{2} C O_{3} = 3.392 \times 10^{- 3} g / 25 m L$
Strength of ${N a}_{2} C O_{3} = 0.1357 g / l i t r e$

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?

25

mL solution containing a mixture of

N a_{2} C O_{3}

N a O H

19.5

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

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

.

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. 1 L solution of

N a_{2} C O_{3}

N a O H

H_{2} O

100

mL of this solution required

20

0.4 M

H C l

in the presence of phenolphthalein. However, another

100

mL sample of the same solution required

25

mL of the same acid in the presence of methyl orange as an indicator.

If molar ratio of

N a_{2} C O_{3}

N a O H

in the original mixture is

\frac{x}{y}

x + y

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.

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Hydrogen Chloride

CHEMISTRY

Watch in App

explore_more_icon

Explore more

Hydrochloric Acid

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon