# Answer to Question #226039 in General Chemistry for pat

Question #226039

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

1
2021-08-16T02:45:32-0400

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.

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