Answer to Question #226039 in General Chemistry for pat

Question

A mass of 3.₃214 g of sodium bi carbonate was dissolved and and diluted to 250,00 ml of aqueous solution .A 20.550X mL of sodium bicarbonate was used to neutralise 25.00mL of H₂SO₄ solution. Calculate the molar concentration of H₂SO₄ SOLUTION

Answer 1

Given: Mass of sodium bi-carbonate i.e. Na₂CO₃ = 3.3214 g

Volume of solution of Na₂CO₃ prepared = 250.00 mL = 0.250 L             (Since 1 L = 1000 mL)

Volume of Na₂CO₃ solution required for neutralization = 20.550 mL

And volume of H₂SO₄ neutralized = 25.00 mL

Molar mass of Na₂CO₃ = Atomic mass of Na X 2 + Atomic mass of C + Atomic mass of O X 3 = 23 X 2 + 12 + 16 X 3 = 106 g/mol.

Since mass = moles X molar mass

=> 3.3214 = moles of Na₂CO₃ X 106

=> Moles of Na₂CO₃ = 0.031334 mol approx.

Since moles of Na₂CO₃ = concentration of Na₂CO₃ X volume of Na₂CO₃ solution prepared in L

=> 0.031334 = Concentration of Na₂CO₃ X 0.250

=> Concentration of Na₂CO₃ = 0.125336 M approx.

The neutralization reaction taking place is given by,

=> Na₂CO₃ + H₂SO₄ --------> Na₂SO₄ + H₂O + CO₂ 

From the above balanced reaction we can see that 1 mole of Na₂CO₃ is required to neutralize 1 mole of H₂SO₄.

Hence moles of Na₂CO₃ required = moles of H₂SO₄ present.

Since moles = concentration X volume of solution

=> Concentration of Na₂CO₃ X volume of Na₂CO₃ solution required = concentration of H₂SO₄ X volume of H₂SO₄ solution neutralized

Hence substituting the values we get,

=> 0.125336 X 20.550 = concentration of H₂SO₄ X 25.00

=> Molar concentration of H₂SO₄ = 0.1030 M approx.