Question

A sample of Na2CO3 weighing 0.53g is added to 101 ml of 0.1 N H2SO4 solution. will the resulting solution be acidic or basic?

Answer 1

According to the question,Here is the Reaction Involved in this Process.

Na₂CO₃ + H₂SO₄ --------→ NA₂SO₄ + CO₂ + 2H₂O

Now, we have to find the Normality of Na₂CO₃ .

Equivalent weight of Na₂CO₃.H₂O is 62 g.

Given weight of Sodium carbonate (Na₂CO₃) = 0.62 g.

Now by Formula,

W = {N . E . V}/{1000}

0.62 = ( N x 62 x 100 ) / 1000

N = 0.62 x 1000 / 62 x 100

N = 620/6200

N= 0.1

Hence the Normality of Na₂CO₃ and H₂SO₄ is Same , 0.1 N .

So the resulting solution will be Neutral.

Answer 2

Answer:

Acidic

Explanation:

Given,

weight of Na₂CO₃ = 0.53g

Equivalent weight of Na₂CO₃ = 53

From normality solution,

1000 mL of Na₂CO₃ of 1 N requires 53 g of Na₂CO₃

100 mL of Na₂CO₃ of 0.1 N requires 0.53 g of Na₂CO₃

For complete neutralization

100 mL of Na₂CO₃ of 0.1 N = 100ml of H₂SO₄ of 0.1 N

We know that,

Equal volume of acid neutralizes equal volume of base, if acid and base have the same normality. So, 100 mL of H₂SO₄, of 0.1 N neutralize 100 mL of Na₂CO₃, of 0.1 N. Thus, 1 ml of acid is left unreacted and the resulting solution must be acidic.