Help With Conquering Chemistry Titration Questions (1 Viewer)

[ikymuk]

Hi If anyone has worked solutions to the Exercises on pg 165 for Conquering Chemistry that would be great I'm having a lot of trouble with the dilutions and wording of each question.

16. 5.267g sodium carbonate was dissolved in water in a volumetric flask and the volume made up to 250ml. 10ml of this solution was pipetted into a conical flask and titrated with hydrochloric acid. 21.3ml was needed to reach the equivalence point. Calculate the molarity of the HCl acid solution.

This solution was then used to determine the concentration of an unknown barium hydroxide solution. 25ml of the barium hydroxide solution required 27.1ml hydrochloric acid solution for exact neutralisation. Calculate the molarity of the barium hydroxide solution. In addition, calculate its concentration in grams per litre.

It's the second part I'm having the most trouble with. Cheers.

 

[RiFiPi]

Data:
5.267g Na2CO3 (with mm = 105.81)
moles = mass/mm -> 5.267/105.81 = 0.0497... moles of Na2CO3
moles = conc*volume (in L) -> 0.0467... = C * 0.25 (250mL) -> 0.1991...mol/L or M

2HCl + Na2CO3 -> H2O + CO2 + 2NaCl
no moles of HCl = 2 * no moles of Na2CO3
C * V (HCl) = 2 * (C * V) (Na2CO3)
C * 0.213 = 2 * (0.1991... * 0.01)
C(HCl) = 0.1869... M

2HCl + Ba(OH)2 -> BaCl2 + 2H2O
no moles of HCl = 2 * no moles Ba(OH)2
0.1869... * 0.0271 = 2 * (C * 0.025)
C[Ba(OH)2] = 0.10 M (to 2 sigfig) = 0.1 mol/L (mm Ba(OH)2 = 171.316)
moles = mass/mm -> 0.1 = mass/171.316 = 17.1316g/L
therefore, 17g/L