Question

1: A solution contains Na2CO3 and NaHCO3. 20 cm of this solution requires 5.0 cm of
0.1M H2SO4 solution for neutralization using phenolphthalein as the indicator.
Methylorange is then added when a further 5.0 cm of 0.2 M H2SO4 was required.
Calculate the masses of Na2CO3 and NaHCO3 in 1 L of this solution.​

Answer 1

Explanation:

2.5 mL of 0.1 M H

2

SO

4

=2.5 mL of 0.2 N H

2

SO

4

=

2

1

Na

2

CO

3

present in 10 mL of mixture

So,

5 mL of 0.2 N H

2

SO

4

=Na

2

CO

3

present in 10 mL of mixture

≡5 mL of 0.2 N Na

2

CO

3

$

≡

1000

0.2×53

×5=0.053 g

Amount of Na

2

CO

3

=

10

0.053

×1000=5.3 g/L of mixture

Between first and second end points,

=2.5 mL of 0.2 M H

2

SO

4

used

=2.5 mL of 0.4 N H

2

SO

4

used

=5 mL of 0.2 N H

2

SO

4

used

≡

2

1

Na

2

CO

3

+NaHCO

3

present in 10 mL of mixture

(5−2.5)mL 0.2 N H

2

SO

4

≡NaHCO

3

present in 10 mL of mixture

≡2.5 mL 0.2 N NaHCO

3

≡

1000

0.2×84

×2.5=0.042 g

Amount of NaHCO

3

=

10

0.042

×1000=4.20g/L of mixture.