The percentages of NaHCO3 and Na2CO3 are to be determined in a mixture of them with KCl. A 0.500 g s — Acid-Base Equilibrium / Titration Chemistry Question
Problem Context
The percentages of NaHCO3 and Na2CO3 are to be determined in a mixture of them with KCl. A 0.500 g sample of the mixture is dissolved in 50.0 mL of deionized water and titrated with 0.115 M HCl, resulting in this pH titration curve.
Write a balanced equation for the reaction that is responsible for the equivalence point that occurs at about
i. pH = 9
Model Answer
CO3 2– + H+ → HCO3 –
ii. pH = 5
Model Answer
HCO3 – + H+ → H2CO3 or HCO3 – + H+ → H2O + CO2
Calculate the total number of moles of acid used to reach each equivalence point if the volumes are 9.63 mL and 34.27 mL, respectively.
Model Answer
9.63 × 10−3 L × 0.115 mol / L = 1.107 × 10−3 mol HCl to titrate CO3 2–
3.427 × 10−3 L × 0.115 mol / L = 3.941 × 10−3 mol HCl to titrate HCO3 –
Determine the number of grams of Na2CO3 and NaHCO3 and their percentages in the original mixture.
Model Answer
1.107 × 10−3 mol CO3 2– × (105.99 g Na2CO3 / mol) = 1.17 × 10–1 g Na2CO3
2.834 × 10−3 mol HCO3 – × (84.01 g NaHCO3 / mol) = 2.38 × 10–1 g NaHCO3
(1.17 × 10–1 g Na2CO3 / 5.00 × 10–1 g mixture) × 100 = 23.4% Na2CO3
(2.38 × 10–1 g NaHCO3 / 5.00 × 10–1 g mixture) × 100 = 47.6% NaHCO3
Sketch a titration curve for a solution of Na2CO3 by itself and describe how it differs from the given curve.
Model Answer
The total volume required to reach the second equivalence point is twice that required to reach the first equivalence point because the number of moles of HCO3 – is equal to the number of moles of CO3 2–.