MyQuestionIcon

1

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

Concentration Terms

20 mL of x ...

Question

$20$ mL of $x$ M $H C l$ neutralises completely $10$ mL of $0.1 M$ $N a H C O_{3}$ and further, $5$ mL of $0.2 M N a_{2} C O_{3}$ solution to methyl orange end-point. The value of $x$ is:

A

0.167 M

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

B

0.133 M

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

C

0.150 M

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

D

0.200 M

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

Open in App

Solution

The correct option is C $0.150 M$
Methyl orange indicator gives the end point corresponding to $100 %$ of sodium carbonate.
The balanced chemical equation is as shown below:
$N a_{2} C O_{3} + 2 H C l \to 2 N a C l + H_{2} O + C O_{2}$
$20 \times x = 10 \times 0.1 + 5 \times 0.2 \times 2$
$x = 0.15 M$
Hence, the value of $x$ is $0.15 M$ .

Suggest Corrections

thumbs-up

0

Similar questions

20

x M

H C l

neutralizes completely

10

0.1 M

N a H C O_{3}

solution and a further

5

0.2 M

N a_{2} C O_{3}

solution to methyl orange end point. The value of

x

Q. Solution "X" contains

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

N a H C O_{3}

. 20 ml of X when titrated using methyl orange indicator consumed 60 ml of 0.1 M

H C l

solution. In another experiment , 20 ml of X solution when titarted using phenolphthalein consumed 20 ml of 0.1 M

H C l

solution. The concentrations (in

m o l {l i t}^{- 1}

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

N a H C O_{3}

in X are respectively :

Q. For a given mixture of

N a H C O_{3}

N a_{2} C O_{3}

, volume of a given

H C l

x

mL with phenolphthalein indicator and further

y

mL required with methyl orange indicator. Hence, volume of

H C l

for complete reaction of

N a H C O_{3}

In the mixture of $N a H C O_{3}$ and $N a_{2} C O_{3}$ volume of a given HCl required is 'x' ml with phenolphthalein indicator and further 'y' ml required with methyl orange indicator. Hence volume of $H C l$ for complete reaction of $N a H C O_{3}$ is

50 m L

of a solution of

N a_{2} C O_{3}

N a H C O_{3}

17.5 m L

0.1 M H_{2} S O_{4}

for neutralisation using phenolphthalein as indicator. Methyl orange is then added when further

25 m L

0.2 M H_{2} S O_{4}

was required. Calculate the weight of

N a_{2} C O_{3}

N a H C O_{3}

respectively in the solution.

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Concentration Terms

CHEMISTRY

Watch in App

explore_more_icon

Explore more

Concentration Terms

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon