1. Quantitative Chemical Analysis Seventh Edition Quantitative Chemical Analysis Seventh Edition Chapter 8-12 Acid-Base Titrations Copyright © 2007 by W. H. Freeman and Company Daniel C. Harris

2. ACTIVITY

3. How to calculate activity? 1: calculate Ionic strength 2: Use Table 8-1

4. BUFFERS

11. = 8.3 H 2 CO 3  H + + HCO 3 - So the lake is a buffer!

13. Consider the titration oftitration 50.00 mL of 0.020 00 M KOH with 0.100 0 M HBr. STEP 2: V e STEP 1: reaction Strong base With Strong acid

14. STEP 3: before equivalence When 3.00 mL of HBr have been added, the reaction is three-tenths complete because Ve = 10.00 mL. Strong base With Strong acid

15. STEP 4: at equivalence pH = 7 Strong base With Strong acid A – 13 B – 12 C – 10 D – 7 E - 1

16. STEP 5: after equivalence The concentration of excess H + at, say, 10.50 mL is given byconcentration Strong base With Strong acid

17. Strong base With Strong acid

18. Weak acid With Strong base EXAMPLE: titrationtitration of 50.00 mL of 0.020 00 M MES (pKa=6.27) with 0.100 0 M NaOH.

19. Weak acid With Strong base HA + OH -  A - + H 2 O STEP 1: reaction STEP 2: V e mols base = mols acid

20. Weak acid With Strong base NB: before any base is added

21. Weak acid With Strong base STEP 3: before equivalence BUFFER! After adding 30% of Ve After adding half of Ve

22. Weak acid With Strong base STEP 4: at equivalence A < 7 B = 7 C > 7

23. Weak acid With Strong base STEP 5: after equivalence excess OH -

24. Weak acid With Strong base Calculated titration curve for the reaction of 50.00 mL of 0.020 00 M MES with 0.100 0 M NaOH. Landmarks occur at half of the equivalence volume (pH = pKa) and at the equivalence point, \ which is the steepest part of the curve.

25. Weak acid With Strong base

26. Weak base With Strong acid

27. 1.435 g sample of dry CaCO 3 and CaCl 2 mixture was dissolved in 25.00 mL of 0.9892 M HCl solution. What was CaCl 2 percentage in original sample, if 21.48 mL of 0.09312 M NaOH was used to titrate excess HCl? During titration 21.48×0.09312=2.000 mmole HCl was neutralized. Initially there was 25.00×0.9892=24.73 mmole of HCl used, so during CaCO 3 dissolution 24.73-2.000=22.73 mmole of acid reacted. As calcium carbonate reacts with hydrochloric acid 1:2 (2 moles of acid per 1 mole of carbonate), original sample contained 22.73/2=11.37 mmole of CaCO 3, or 1.137 g (assuming molar mass of CaCO 3 is 100.0 g). So original sample contained 1.137/1.435×100%=79.27% CaCO 3 and 100.0-72.27%=20.73% CaCl 2. Back titration