Principles of Chemistry II © Vanden BoutTodayTitrationdetermining something about an unknownby reacting it with a known solutionNeutralization (again)we'll need this to figure out titrationSolubilityThe easiest of all the equilibriawe'll need this for polyprotic acids Principles of Chemistry II © Vanden BoutLast Time Strong Acid/Strong Base Titrationat the equivalence point we have equal number of moles of acid and baseoriginal solution 50 mL HCladding .1 M NaOHat equivalence pointsame number of moles of base.1L x .1M = 0.01 moles OH-therefore the solution originallyhad 0.01 moles H+concentration was .2 M Principles 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 1.0412.86 Principles of Chemistry II © Vanden BoutStrong Acid/Strong Base TitrationStrong AcidNeutralStrong BasePrinciples of Chemistry II © Vanden BoutWhat volume of a 1 M NaOH will you need to add to 200 mL of a 0.2 M solution of HCl to neutralize it?! A.!! 10 mL ! B.!! 20 mL ! C.!! 30 mL!! D.!! 40 mL! E.!! 400 mLThere are .04 moles of H+ .2M x .2Lto neutralize you'll need .04 moles of OH- For that you'll need .04L of a 1M solution Principles 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 MAt the endpoint there are equal moles H+ and OH-0.04 L x 1 M = 0.04 moles OH-.04 moles H+/0.2L = 0.2 M Principles 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< 1 Principles of Chemistry II © Vanden Boutcolor just barely changing forPhenolpthaleinPrinciples of Chemistry II © Vanden BoutBromophenol Blue has a pKa of around 4. When it is protonated it is green, when it is deprotonated it is blue. What color would in be in a solution in which the pH was 8?! A.!! blue ! B.!! green ! C.!! a mix of blue and greenpKa = 4 Ka = 10-4pH = 8 [H+] = 10-8[H+]<<Ka [A-]/[HA]>>1all deprotonatedblue Principles of Chemistry II © Vanden BoutTitration with weak acid/baseAll the sameNeutralize firstThen equilibriumNeutralization reactionsH+(aq) + OH-(aq) H2O(l)OH-(aq) + HA(aq) A-(aq)OH-(aq) + BH+(aq) B(aq)H+(aq) + A-(aq) HA(aq)H+(aq) + B(aq) BH+(aq) Principles of Chemistry II © Vanden BoutI have a 100 mL of a 1 M solution of acetic acidI add 100 mL of 0.5 M NaOHWhat remains in the solution?! A.!! 0.1 moles of acetic acid ! B.!! 0.1 moles acetic acid and 0.05 moles of acetate ! C.!! 0.05 moles of acetic acid and 0.05 moles of acetate! D.!! 0.05 moles of acetic acid and 0.1 moles of acetate! E.!! 0.1 moles of acetate.1L x 1 M = 0.1 moles acetic acid.1L x .5M = 0.05 moles of OH-neutralize OH- and HAleft with .05 moles of HAand 0.05 moles of A- Principles of Chemistry II © Vanden BoutIf I have a solution that has 0.05 moles of acetic acid and 0.05 moles of acetate what do I have?! A.!! strong acid solution ! B.!! weak acid solution ! C.!! buffer! D.!! weak base! E.!! strong basePrinciples of Chemistry II © Vanden BoutOH-(aq) + HA(aq) A-(aq)H+(aq) + A-(aq) HA(aq)Neutralization of a weak acid or weak base will yield a bufferbecause you generatethe conjugate base or acidBuffer will remain until you react all of the initial acid or base Principles of Chemistry II © Vanden BoutTitrating a weak acidbuffer regionhalf-way to equivalence pointhalf of HA converted to A-[HA] = [A-][H+] = KapH = pKaEquivalence pointAll of HA converted to A-Now a weak base solutionall HA neutralized at equivalence pointAdding OH- now makes a strong base solution Principles of Chemistry II © Vanden BoutCalc on Doc Cam Principles of Chemistry II © Vanden BoutAll have the same concentration so they all have the same equivalence point (end point)Principles of Chemistry II © Vanden BoutWeak base titrated with strong acid Principles of Chemistry II © Vanden BoutRolaids® contain about 0.1 g of Magnesium HydroxideWhy in the world would you ever put such a thing in your mouth? ! A.!! 0.1 g is nothing. I each 10-20 g NaOH daily just for laughs! B.!! Acids are dangerous by bases as quite safe ! C.!! The saliva in my mouth is acidic enough to "handle it"! D.!! Mg(OH)2 is not soluble in water Principles of Chemistry II © Vanden BoutSolubility EquilibriaMg(OH)2 (s) Mg2+(aq) + 2OH-(aq)K = [Mg2+][OH-]2spsolubility product Principles of Chemistry II © Vanden
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