Answer to Question #107922 in Organic Chemistry for Abdul

Question

A sample containing 25.14 g of neutral salts, glucose and a sodium carbonate/bicarbonate
buffer was dissolved in 100 ml of water. A 25 ml aliquot of the resultant solution required
20.35 ml of 0.0987 M HCl when titrated to the PP end-point. A second 25 ml aliquot was
titrated to the MO end-point and required 56.75 ml of the acid. Calculate the percentage of
Na2CO3 (molecular weight 106) and NaHCO3 (molecular weight 84) in the sample

Answer 1

Na₂CO₃ + HCl = NaHCO₃ + NaCl

NaHCO₃ + HCl = H₂O + CO₂ + NaCl

m(Na₂CO₃) = 2 × (0.0987 × 20.35 / 1000) × 4 × 106 = 1.70 g

m(NaHCO₃) = ((56.75 - 2 × 20.35) / 1000) × 0.0987 × 4 × 84 = 0.53 g

$\omega$ (Na₂CO₃) = (1.70 / 25.14) × 100 = 6.76 %

$\omega$ (NaHCO₃) = (0.53 / 25.14) × 100 = 2.11 %