Principles of Chemistry II © Vanden BoutTo d ayTitrationdetermining something about an unknownby reacting it with a known solutionPolyprotic AcidsPrinciples of Chemistry II © Vanden BoutTitrationWhy do a titration.You have a solution with an unknown propertyUnknown Concentration?Unknown Ka (Kb)?Both.Slowly neutralize the solution by addinga strong base (acid)monitor the pH with each additionPrinciples of Chemistry II © Vanden BoutNeutralize firstThen look at the neutralization from last class equilibriumimagine a 100 mL solution with 0.1 M Acetic Acid (initial .01 moles)we add 10 mL of 0.1M NaOH in each titration step (add 0.001 moles)HA OH-Initial (moles)After Neutralization EquilibriumpHVolume (L)0.0100.0000.0100.102.870.113.79A-HA OH-A-0.0010.000.000.0100.0000.009 0.0000.0000.001Principles of Chemistry II © Vanden BoutNeutralize firstThen look at the neutralization from last class equilibriumimagine a 100 mL solution with 0.1 M Acetic Acid (initial .01 moles)we add 10 mL of 0.1M NaOH in each titration step (add 0.001 moles)HA OH-Initial (moles)After Neutralization EquilibriumpHVolume (L)0.0100.0000.0100.102.870.113.790.009 0.124.15A-HA OH-A-0.0010.0010.000.000.000.0100.0000.0090.0080.0000.0000.0000.0010.002Principles of Chemistry II © Vanden BoutNeutralize firstThen look at the neutralization from last class equilibriumimagine a 100 mL solution with 0.1 M Acetic Acid (initial .01 moles)we add 10 mL of 0.1M NaOH in each titration step (add 0.001 moles)HA OH-Initial (moles)After Neutralization EquilibriumpHVolume (L)0.0100.0000.0100.10.........2.870.113.790.009 0.124.15A-HA OH-A-0.0010.0010.000.000.000.0100.0000.0090.0080.0000.0000.0000.0010.0020.006 0.001 0.0050.0050.000 0.005 0.15 4.75Principles of Chemistry II © Vanden BoutNeutralize firstThen look at the neutralization from last class equilibriumimagine a 100 mL solution with 0.1 M Acetic Acid (initial .01 moles)we add 10 mL of 0.1M NaOH in each titration step (add 0.001 moles)HA OH-Initial (moles)After Neutralization EquilibriumpHVolume (L)0.0100.0000.0100.10.........2.870.113.790.009 0.124.15A-HA OH-A-0.0010.0010.000.000.000.0100.0000.0090.0080.0000.0000.0000.0010.0020.006 0.001 0.0050.0050.000 0.005 0.15.........0.001 0.001 0.0090.0000.000 0.010 0.154.758.78Principles of Chemistry II © Vanden BoutNeutralize firstThen look at the neutralization from last class equilibriumimagine a 100 mL solution with 0.1 M Acetic Acid (initial .01 moles)we add 10 mL of 0.1M NaOH in each titration step (add 0.001 moles)HA OH-Initial (moles)After Neutralization EquilibriumpHVolume (L)0.0100.0000.0100.10.........2.870.113.790.009 0.124.15A-HA OH-A-0.0010.0010.000.000.000.0100.0000.0090.0080.0000.0000.0000.0010.0020.006 0.001 0.0050.0050.000 0.005 0.15.........0.001 0.001 0.0090.0000.000 0.010 0.150.000 0.001 0.0100.0000.001 0.010 0.164.758.7812.8Principles of Chemistry II © Vanden BoutTitrating a weak acidPrinciples of Chemistry II © Vanden BoutTitrating a weak acidPrinciples of Chemistry II © Vanden BoutTitrating a weak acidPrinciples of Chemistry II © Vanden BoutEasier Version Strong Acid/Strong Baseat the equivalence point we have equal number of moles of acid and baseNo Buffer(conjugate base is infinitely weak)HCl titrated with NaOHPrinciples of Chemistry II © Vanden BoutNeutralize firstThen look at the equilibriumimagine a 100 mL solution with 0.1 moles of HCl we add .01 moles of NaOH in each titration step (10 mL of 1M)mol H+mol OH-Initial After Neutralizationmol H+mol OH-EquilibriumpHVolume (L)pOH0.1 0.010.0 0.010.09 0.010.09 0.000.01 0.01 0.00 0.000.11.........0.09 13.910.08 0.00 0.12 0.18 13.820.08 0.01 0.07 0.00 0.13 0.27 13.760.02 0.01 0.01 0.00 0.19 12.721.280.20 7.007.000.0 0.01 0.21 1.3312.670.0 0.02 0.0 0.02 0.22 1.0412.86Principles of Chemistry II © Vanden BoutStrong Acid/Strong Base TitrationStrong AcidNeutralStrong BasePrinciples of Chemistry II © Vanden BoutAt the endpoint of your titration you have added 40 mL of a 1M NaOH solution to 200 mL of an unknown HCl solution. What was the concentration of the HCl? ! A.! ! 0.1 M ! B.! ! 0.2 M ! C.! ! 0.4 M!! D.! ! 1 M! E.! ! 2 MPrinciples of Chemistry II © Vanden BoutThe equivalence point in this titration is at ! A. 5 mL! B. 14 mL ! C. 28 mL!! D. 44 mLPrinciples of Chemistry II © Vanden BoutBelow is a titration curve of a ___________! A. weak acid with a strong base ! B. weak base with a strong acid ! C. strong acid with a strong base!! D. strong base with a strong acidPrinciples of Chemistry II © Vanden BoutThe half-equivalence point in this titration is at ! A. 5 mL! B. 14 mL ! C. 28 mL!! D. 44 mLPrinciples of Chemistry II © Vanden BoutThe pKa of this weak acid is_______! A. 3.1 ! B. 4.7 !C. 8!! D. 12.1Principles of Chemistry II © Vanden BoutFinding the endpoint (equivalence point)Indicator dyepKa = 8.2 Ka = 6.3 x 10-9Ka = [H+] x[HA]Phenolphthalein[A-]amount of indicator is so small it doesn't affect the pH, but the equilibrium of the dye is strongly affected by the pHHAA-= [H+] xClearPink[H+] < 6.3 x 10-9 pH>8.2ClearPink> 1 [H+] > 6.3 x 10-9 pH < 8.2ClearPink< 1Principles of Chemistry II © Vanden Boutcolor just barely changing forPhenolpthaleinPrinciples of Chemistry II © Vanden BoutBromophenol Blue has a pKa of
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