Question

Mixture of na2co3 and nahco3 has a mass of 22 g. treatment with excess hci solution liberates 6 l of co2 at 25°c and 0.947 atm pressure. determine the percent na2co3 in the mixture.

Answer 1

Answer:

When phenolphthalein is the indicator, whole of NaOH has been neutralised and carbonate converted into bicarbonate, i.e.,

NaOH+HCl→NaCl+H

2

O

Na

2

CO

3

+HCl→NaHCO

3

+NaCl

So, 25 mL

10

N

HCl≡NaOH+1/2Na

2

CO

3

present in 25 mL of mixture.

In another titration, when methyl orange is the indicator, whole of NaOH has been neutralised and carbonate converted into carbonic acid, i.e.,

Na

2

CO

2

+2HCl→2NaCl+H

2

CO

3

30mL

10

N

HCl≡NaOH+Na

2

CO

3

present in 25 mL of mixture

Hence,

(30−25)mL

10

N

HCl≡

2

1

Na

2

CO

3

present in 25 mL of mixture

Hence,

10 mL

10

N

HCl≡Na

2

CO

3

present in 25 mL of mixture

≡10 mL

10

N

Na

2

CO

3

solution

Amount of Na

2

CO

3

=

10×1000

53×10

=0.053 g

This amount of Na

2

CO

3

is present in 25 mL of mixture.

This amount present in one litre of mixture,

=

25

0.053

×1000=2.12 g

(30−10)mL

10

N

HCl≡NaOH present in 25 mL of mixture

≡20 mL

10

N

NaOH

Amount of NaOH in 25 mL of mixture =

10×1000

40×20

=0.08 g

The amount present in one litre of mixture =

25

0.08

×1000=3.20 g.