INTRODUCTION TO TITRIMETRY 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 TITRATIONS IN PRACTICE 1 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 2 Slowly add standard solution from a burette to the sample solution Known: concentration of the titrant 3 Add until just enough titrant is added to react with all the analyte The end point is signalled by some physical change or detected by an instrument Note the volume of titrant used Known: volume of the titrant 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 cHA vHA cBOH vBOH 1 1 cHA c1v1 c2 v2 n1 n2 cBOH vBOH vHA OR if we have: H2A + 2BOH 2BA + 2H2O analyte titrant Then from the balanced equation we know: 1 mol H2A reacts with 2 mol BOH We also know: CBOH, VBOH and VHA and c1v1 c2 v2 n1 n2 c BOH v BOH c HA v HA c BOH v BOH c HA 2vHA 1 2 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. 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 EQUIVALENCE POINT The amount of added titrant is the exact amount necessary for stoichiometric reaction with the analyte in the sample. VS END POINT 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: Et = Veq – Vep Estimated with a blank titration Indicators used to observe the end point (at/near the equivalence point) Thymol blue indicator 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. BACK TITRATION Add excess titrant and then determine the excess amount of unreacted titrant 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 For example: If I add excess titrant … H2A + 2BOH 2BA + 2H2O analyte titrant reacted Known: vanalyte, cBOH and vBOH(total) Unknown: vBOH(reacted) … and then react the excess with a second titrant as follows: HX + BOH BA + H2O titrant 2 excess Known: vHX, cHX and cBOH Work backwards! Step 1: From the second titration HX + BOH BA + H2O titrant 2 Known: vHX, cHX and cBOH excess cHX vHX vBOH (excess) cBOH Step 2: Calculate Unknown vBOH(reacted) Known: vBOH(total) vBOH(reacted) = vBOH(total) – vBOH(excess) Step 3: From the first titration H2A + 2BOH 2BA + 2H2O analyte cH2A titrant reacted cBOHvBOH (reacted) 2vH2A Known: vanalyte, cBOH and vBOH(reacted) 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? 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 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.