Question

A sample of Na2CO3.H2O weighing 0.62 g is added to 100 ml of 0.1 N H2SO4. Will the resulting solution be neutral? If yes then how?

Answer 1

Hello Dear.

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 =  NEV/100

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.

Hope it Helps. :-)

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Answer 2

$\underline{\underline{\sf{\color{orange}{QUESTION:-}}}}$

◘ A sample of Na2CO3.H2O weighing 0.62 g is added to 100 ml of 0.1 N H2SO4. Will the resulting solution be neutral? If yes then how?

$\underline{\underline{\sf{\color{orange}{SOLUTION:-}}}}$

◘ When we add Sodium Carbonate (Na₂CO₃) to Sulphuric Acid (H₂SO₄), the reaction is as follows :-

Na₂CO₃. H₂O + H₂SO₄ $\longrightarrow$ Na₂SO₄ + CO₂ + H₂O

Calculating the normality of Sodium Carbonate :

• Given weight of Sodium Carbonate (W) = 0.62 g
• Equivalent weight of Sodium Carbonate (E) = 62 g

We know,

$\sf{W=\dfrac{N\times E \times V}{1000} }$

Calculating further :

$\longmapsto \displaystyle \sf{ 0.62=\frac{N\times62\times100}{1000} }$

$\longmapsto \displaystyle \sf{ 620=N\times62\times100}$

$\longmapsto \displaystyle \sf{ 10N=1}$

$\longmapsto \displaystyle \sf{ N=\frac{1}{10} }$

$\longmapsto \displaystyle \sf{ N=0.1}$

Normality of Sodium Carbonate and Sulphuric Acid is same. So, the resulting solution will be neutral.