A 3.75 g sample of baking soda containing Na2CO3 and NaHCO3, and inert material

Question

A 3.75 g sample of baking soda containing Na2CO3 and NaHCO3, and inert material was dissolved in water and diluted to 500 mL. A 50 mL aliquot of this solution required 46.93 mL of 0.1208 M HCl to reach the bromocresol green end point. Another 50 mL aliquot was treated with 10 mL of 0.2506 M NaOH to convert the bicarbonate to carbonate ion, which then was precipitated by the addition of excess BaCl2. After removal of the precipitated BaCO3, the excess NaOH in the solution required 14.09 mL of 0.09783 M HCl for titration. Calculate % Na2CO3 and % NaHCO3 in the sample.

Explanation / Answer

total mass of material = 3.75 g

For 50 ml aliquot,

Total HCl required for neutralization = 0.1208 M x 46.93 ml = 5.6691 mmol

So the total Na2CO3 + NaHCO3 present = 5.6691 mmol

moles of Na2CO3 present = 5.6691 x 2/3 = 3.7794

for another 50 ml aliquot,

moles of NaOH added = 0.2506 M x 10 ml = 2.506 mmol

moles of HCl reacted with excess NaOH = 0.09783 M x 14.09 ml = 1.378425 mmol

moles of NaOH reacted = 2.506 - 1.378425 = 1.1276 mmol

moles of NaHCO3 present in 50 ml aliquot = 1.1276 mmol

mass of NaHCO3 = 1.1276 mmol x 84 x 10/1000 = 0.9472 g

% NaHCO3 in the sample = 0.9472 x 100/3.75 = 25.26%

moles of Na2CO3 present = 5.6691 - 1.1276 = 4.5415 mmol

mass of Na2CO3 = 4.5415 mmol x 10 x 106/2 x 1000 = 2.407 g

% Na2CO3 in the sample = 2.407 x 100/3.75 = 64.20%

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