1. INTRODUCTION TO TITRIMETRY

2. Most common types of titrations : acid-base titrations oxidation-reduction titrations complex formation precipitation reactions In a titration, increments of titrant are added to the analyte until their reaction is complete. From the quantity of titrant required, the quantity of analyte that was present can be calculated.

3. TITRATIONS IN PRACTICE Accurately add of specific volume of sample solution to a conical flask using a pipette Known: volume of sample Unknown: concentration of analyte in sample 1

4. Slowly add standard solution from a burette to the sample solution Known: concentration of the titrant 2

5. Add until just enough titrant is added to react with all the analyte The end point is signaled by some physical change or detected by an instrument Known: volume of the titrant Note the volume of titrant used 3

6. If we have: HA + BOH  BA + H 2 O Then from the balanced equation we know: 1 mol HA reacts with 1 mol BOH analytetitrant We also know: C BOH, V BOH and V HA and 

7. STANDARD SOLUTIONS Standard solution: Reagent of known concentration Primary standard: highly purified compound that serves as a reference material in a titration. Determine concentration by dissolving an accurately weighed amount in a suitable solvent of known volume.

8. Secondary standard: compound that does not have a high purity Determine concentration by standardisation. Titrate standard using another standard. Standard solutions should: Be stable React rapidly with the analyte React completely with the analyte React selectively with the analyte

9. EQUIVALENCE POINT The amount of added titrant is the exact amount necessary for stoichiometric reaction with the analyte in the sample. An estimate of the equivalence point that is observed by some physical change associated with conditions of the equivalence point. Aim to get the difference between the equivalence point and the end point as small as possible. Titration error:E t = V eq – V ep Estimated with a blank titration END POINTVS

10. Indicators  used to observe the end point (at/near the equivalence point) Thymol blue indicator

11. Instruments can also be used to detect end points. Respond to certain properties of the solution that change in a characteristic way. E.g.: voltmeters, ammeters, ohmmeters, colorimeters, temperature recorders, refractometers etc.

12. Add excess titrant and then determine the excess amount unreacted by back titration with a second titrant. BACK TITRATION Used when: end point of back titration is clearer than end point of direct titration an excess of the first titrant is required to complete reaction with the analyte

13. If we have: HA + BOH  BA + H 2 O Then from the balanced equation we know: 1 mol HA reacts with 1 mol BOH analytetitrant If I add excess titrant and then react the excess with a second titrant as follows: HX + BOH  BA + H 2 O titrant 2excess Then from the balanced equation we know: 1 mol HX reacts with 1 mol BOH

14. We also know: C BOH, C HX and V HX and  We also know v BOH (total)  v BOH (reacted) = v BOH (total) – v BOH (excess)

15.  From our initial titration: HA + BOH  BA + H 2 O analyte titrant we then know: C BOH, V BOH (reacted) and V HA and we want to find C HA !

16. In summary: HA + BOH  BA + H 2 O analytetitrant reacted V HA C BOH HX + BOH  BA + H 2 O titrant 2titrant excess V HX C HX C BOH  v BOH (reacted) = v BOH (total) – v BOH (excess) C HA ?

17. Example: 50.00 ml of HCl was titrated with 0.01963M Ba(OH) 2. The end point was reached (using bromocresol green as indicator) after 29.71 ml Ba(OH) 2 was added. What is the concentration of the HCl? 2HCl + Ba(OH) 2  BaCl 2 + 2H 2 O 50.00 ml 29.71 ml 0.01963M  C 1 = 0.02333 M = [HCl]

18. Example: A 0.8040 g sample of iron ore is dissolve in acid. The iron is reduced to Fe 2+ and titrated with 0.02242 M KMnO 4. 47.22 ml of titrant was added to reach the end point. Calculate the % Fe in the sample. MnO 4 - + 5Fe 2+ + 8H +  Mn 2+ + 5Fe 3+ + 4H 2 O 0.02242 M 47.22 ml n = cv n = (0.02242 M)(0.04722 L) n = 1.059x10 -3 mol BUT n Fe2+ =

19. MnO 4 - + 5Fe 2+ ….. 0.02242 M 47.22 ml 1.059x10 -3 mol 5.293x10 -3 mol M fe = 55.847 g/mol m = (5.293x10 -3 mol)(55.847 g/mol) m = 0.2956 g Fe in sample

20. Example: The CO in a 20.3 L sample of gas was converted to CO 2 by passing the gas over iodine pentoxide heated to 150 o C: I 2 O 5 (s) + 5CO(g)  5CO 2 (g) + I 2 (g) The iodine distilled at this temperature was collected in an absorber containing 8.25 mL of 0.01101 M Na 2 S 2 O 3 : I 2 (aq) + 2S 2 O 3 2- (aq)  2I - (aq) + S 4 O 6 2- (aq) The excess Na 2 S 2 O 3 was back titrated with 2.16 mL of 0.00947 M I 2 solution. Calculate the mg CO per liter of sample.

21. I 2 O 5 (s) + 5CO(g)  5CO 2 (g) + I 2 (g) I 2 (aq) + 2S 2 O 3 2- (aq)  2I - (aq) + S 4 O 6 2- (aq) 8.25 mL 0.01101 M added I 2 (aq) + 2S 2 O 3 2- (aq)  2I - (aq) + S 4 O 6 2- (aq) 2.16 mL 0.00947 M  n = 2.05x10 -5 mol  n added = 9.08x10 -5 mol  n reacted = n added - n excess = 4.99x10 -5 mol  n excess = 2(2.05x10 -5 ) mol n excess = 4.09x10 -5 mol ADDED EXCESS REACTED

22. I 2 (aq) + 2S 2 O 3 2- (aq)  2I - (aq) + S 4 O 6 2- (aq) n reacted = 4.99x10 -5 mol 2n I2 produced = n reacted n I2 = 2.50x10 -5 mol I 2 O 5 (s) + 5CO(g)  5CO 2 (g) + I 2 (g) I 2 (aq) + 2S 2 O 3 2- (aq)  2I - (aq) + S 4 O 6 2- (aq) I 2 O 5 (s) + 5CO(g)  5CO 2 (g) + I 2 (g) n CO = 5n I2 n CO = 1.25x10 -4 mol n I2 produced = 2.50x10 -5 mol

23. n CO = 1.25x10 -4 mol Calculate the mg CO per liter of sample. M CO = 28.01g/mol  m CO = 3.49x10 -3 g V sample = 20.3 L  3.49 mg / 20.3 L = 0.172 mg/L CO in the sample