Answer to Question #128221 in General Chemistry for tasnim

Question

10 g of a mixture of sodium chloride and anhydrous sodium carbonate were made up to 1 dm3 of aqueous solution. 25cm3 of this solution required 20 cm3 of 0.2 M hydrochloric acid for neutralization.What was the mass of sodium chloride in the mixture

Answer 1

25 cm³ of this solution required 20 cm³ of 0.2 M ( = 0.004 mol) hydrochloric acid for neutralization. Therefore, 1 dm³ of solution required 1000 * 0.004 / 25 = 0.16 mol of hydrochloric acid. It means that there is 0.16 / 2 = 0.08 mol of Na₂CO₃ in 10 g of a mixture. It is equivalent to 0.08 * 106 = 8.48 g of compound. This means that the remaining NaCl weighs 10 - 8.48 = 1.52 g.