Question #226039

A mass of 3._{3}214 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 H_{2}SO_{4 }solution. Calculate the molar concentration of H_{2}SO_{4 }SOLUTION

Expert's answer

Given: Mass of sodium bi-carbonate i.e. Na_{2}CO_{3}Â = 3.3214 g

Volume of solution of Na_{2}CO_{3}Â prepared = 250.00 mL = 0.250 LÂ Â Â Â Â Â Â Â Â Â Â Â Â (Since 1 L = 1000 mL)

Volume of Na_{2}CO_{3}Â solution required for neutralization = 20.550 mL

And volume of H_{2}SO_{4}Â neutralized = 25.00 mL

Molar mass of Na_{2}CO_{3}Â = 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 Na_{2}CO_{3}Â X 106

=> Moles of Na_{2}CO_{3}Â = 0.031334 molÂ approx.

Since moles of Na_{2}CO_{3}Â = concentration of Na_{2}CO_{3}Â X volume of Na_{2}CO_{3}Â solution prepared in L

=> 0.031334 = Concentration of Na_{2}CO_{3}Â X 0.250

=> Concentration of Na_{2}CO_{3}Â = 0.125336 M approx.

The neutralization reaction taking place is given by,

=> Na_{2}CO_{3}Â + H_{2}SO_{4}Â --------> Na_{2}SO_{4}Â + H_{2}O + CO_{2}Â

From the above balanced reaction we can see that 1 mole of Na_{2}CO_{3}Â is required to neutralize 1 mole of H_{2}SO_{4}.

Hence moles of Na_{2}CO_{3}Â required = moles of H_{2}SO_{4}Â present.

Since moles = concentration X volume of solution

=> Concentration of Na_{2}CO_{3}Â X volume of Na_{2}CO_{3}Â solution required = concentration of H_{2}SO_{4}Â X volume of H_{2}SO_{4}Â solution neutralized

Hence substituting the values we get,

=> 0.125336 X 20.550 = concentration of H_{2}SO_{4}Â X 25.00

**=> Molar concentration of H**_{2}**SO**_{4}**Â = 0.1030 M approx.**

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