MyQuestionIcon

1

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

Neutralisation Reaction

A solution co...

Question

A solution contains $N a_{2} C O_{3}$ and $N a H C O_{3}$ . $10 m L$ of the solution required $2.5 m L$ of $0.1 M H_{2} S O_{4}$ for neutralisation using phenolphthalein as indicator. Methyl orange is then added when a further $2.5 m L$ of $0.2 M H_{2} S O_{4}$ was required. Calculate the amount of $N a_{2} C O_{3}$ and $N a H C O_{3}$ in one litre of the solution.

Open in App

Solution

$2.5 m L$ of $0.1 M H_{2} S O_{4} = 2.5 m L$ of $0.2 N H_{2} S O_{4}$
$= \frac{1}{2} N a_{2} C O_{3}$ present in $10 m L$ of mixture
So,
$5 m L$ of $0.2 N H_{2} S O_{4} = N a_{2} C O_{3}$ present in $10 m L$ of mixture
$\equiv 5 m L$ of $0.2 N N a_{2} C O_{3}$ $
$\equiv \frac{0.2 \times 53}{1000} \times 5 = 0.053 g$
Amount of $N a_{2} C O_{3} = \frac{0.053}{10} \times 1000 = 5.3 g / L$ of mixture
Between first and second end points,
$= 2.5 m L$ of $0.2 M H_{2} S O_{4}$ used
$= 2.5 m L$ of $0.4 N H_{2} S O_{4}$ used
$= 5 m L$ of $0.2 N H_{2} S O_{4}$ used
$\equiv \frac{1}{2} N a_{2} C O_{3} + N a H C O_{3}$ present in $10 m L$ of mixture
$(5 - 2.5) m L 0.2 N H_{2} S O_{4}$
$\equiv N a H C O_{3}$ present in $10 m L$ of mixture
$\equiv 2.5 m L 0.2 N N a H C O_{3}$
$\equiv \frac{0.2 \times 84}{1000} \times 2.5 = 0.042 g$
Amount of $N a H C O_{3} = \frac{0.042}{10} \times 1000 = 4.20 g / L$ of mixture.

Suggest Corrections

thumbs-up

1

Similar questions

Q. A solution contains

N a_{2} C O_{3} a n d N a H C O_{3}

.10 ml of the solution requires 2.5 ml of 0.1M

H_{2} S O_{4}

for neutralisation using phenolphthalein as an indicator. Methyl orange is added when further 2.5 ml of 0.2 M

H_{2} S O_{4}

is required. The amount of

N a H C O_{3}

in one litre of the solution 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.

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.

Q. A solution contain

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

N a H C O_{3}

. 10 ml of this solution required 2.5 ml of 0.1 M

H_{2} {S O}_{4}

for neutralization using phenolphthalein as indicator. Methyl orange is then added when a further 2.5 ml of 0.2 M

H_{2} {S O}_{4}

was required. The amount of

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

N a H C O_{3}

in 1 litre of the solution is ?

Q. A solution contained

N a_{2} C O_{3}

N a H C O_{3}

15 m l

of the solution required

5 m l

\frac{N}{10} H C l

for neutralization using phenolphthalein as an indicator. Addition of methyl orange required a further

15 m l

of the acid for neutralization. The amount of

N a_{2} C O_{3}

present in the solution is:

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Acid Base Reactions

CHEMISTRY

Watch in App

explore_more_icon

Explore more

Neutralisation Reaction

Standard X Chemistry

Join BYJU'S Learning Program

CrossIcon