INTRODUCTION TO TITRIMETRY

1. INTRODUCTION TO TITRIMETRY

2. 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. • Most common types of titrations : • acid-base titrations • oxidation-reduction titrations • complex formation • precipitation reactions

3. 1 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

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

5. 3 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 Note the volume of titrant used Known: volume of the titrant

6. If we have: HA + BOH  BA + H2O analyte titrant Then from the balanced equation we know: 1 mol HA reacts with 1 mol BOH We also know: CBOH, VBOH and VHA 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. VS END POINT EQUIVALENCE POINT An estimate of the equivalence point that is observed by some physical change associated with conditions of the equivalence point. The amount of added titrant is the exact amount necessary for stoichiometric reaction with the analyte in the sample. Aim to get the difference between the equivalence point and the end point as small as possible. Titration error: Et = Veq – Vep Estimated with a blank titration

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. BACK TITRATION Add excess titrant and then determine the excess amount unreacted by back titration with a second titrant. • 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 + H2O analyte titrant Then from the balanced equation we know: 1 mol HA reacts with 1 mol BOH If I add excess titrant and then react the excess with a second titrant as follows: HX + BOH  BA + H2O titrant 2 excess Then from the balanced equation we know: 1 mol HX reacts with 1 mol BOH

14. We also know: CBOH, CHX and VHX and  We also know vBOH(total)  vBOH(reacted) = vBOH(total) – vBOH(excess)

15. From our initial titration: HA + BOH  BA + H2O analyte titrant we then know: CBOH, VBOH(reacted) and VHA and we want to find CHA! 

16. In summary: HA + BOH  BA + H2O analyte titrant reacted VHA CBOH CHA? HX + BOH  BA + H2O titrant 2 titrant excess CHX CBOH VHX  vBOH(reacted) = vBOH(total) – vBOH(excess)

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 BaCl2 + 2H2O 50.00 ml 29.71 ml 0.01963M C1 = 0.02333 M = [HCl]

18. Example: A 0.8040 g sample of iron ore is dissolve in acid. The iron is reduced to Fe2+ and titrated with 0.02242 M KMnO4. 47.22 ml of titrant was added to reach the end point. Calculate the % Fe in the sample. MnO4- + 5Fe2+ + 8H+ Mn2+ + 5Fe3+ + 4H2O 0.02242 M 47.22 ml BUT nFe2+= n = cv n = (0.02242 M)(0.04722 L) n = 1.059x10-3 mol

19. MnO4- + 5Fe2+….. 5.293x10-3 mol 0.02242 M 47.22 ml 1.059x10-3 mol Mfe = 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 CO2 by passing the gas over iodine pentoxide heated to 150oC: I2O5(s) + 5CO(g)  5CO2(g) + I2(g) The iodine distilled at this temperature was collected in an absorber containing 8.25 mL of 0.01101 M Na2S2O3: I2(aq) + 2S2O32-(aq)  2I-(aq) + S4O62-(aq) The excess Na2S2O3 was back titrated with 2.16 mL of 0.00947 M I2 solution. Calculate the mg CO per liter of sample.

21. 8.25 mL 0.01101 M added I2O5(s) + 5CO(g)  5CO2(g) + I2(g) I2(aq) + 2S2O32-(aq)  2I-(aq) + S4O62-(aq) ADDED nadded = 9.08x10-5 mol I2(aq) + 2S2O32-(aq)  2I-(aq) + S4O62-(aq) EXCESS 2.16 mL 0.00947 M nexcess = 2(2.05x10-5) mol nexcess = 4.09x10-5 mol n = 2.05x10-5 mol nreacted =nadded-nexcess= 4.99x10-5 mol REACTED

22. I2O5(s) + 5CO(g)  5CO2(g) + I2(g) I2(aq) + 2S2O32-(aq)  2I-(aq) + S4O62-(aq) I2(aq) + 2S2O32-(aq)  2I-(aq) + S4O62-(aq) nreacted = 4.99x10-5 mol 2nI2 produced = nreacted nI2 produced = 2.50x10-5 mol I2O5(s) + 5CO(g)  5CO2(g) + I2(g) nI2 = 2.50x10-5 mol nCO = 5nI2 nCO = 1.25x10-4 mol

23. Calculate the mg CO per liter of sample. nCO = 1.25x10-4 mol MCO = 28.01g/mol mCO = 3.49x10-3 g Vsample = 20.3 L 3.49 mg / 20.3 L = 0.172 mg/L CO in the sample