Principles of Chemistry II © Vanden BoutHA(aq) H+(aq) + A-(aq)How are we going to control this equilibrium?Add HA shift to the "products"Add A- shift to the "reactants"Add H+ shift to the "reactants"Remove H+ shift to the "products"Principles of Chemistry II © Vanden BoutNeutralizationA solution can be neutralized (equal amounts of H+ and OH-) by adding an acid or base to the solutionAs you are mixing two solutions, it is generally easiest to think in terms of moles (rather than molarity)Principles of Chemistry II © Vanden BoutWhat volume of a 0.1 M NaOH will you need to add to 200 mL of a 0.2 M solution of HCl to neutralize it?! A.! ! 100 mL ! B.! ! 200 mL ! C.! ! 300 mL!! D.! ! 400 mL! E.! ! 500 mLThere are .04 moles of H+ .2M x .2Lto neutralize you'll need .04 moles of OH- For that you'll need .4L of a .1M solutionOr you can look at it as the acid is twiceas concentrated as the basetherefore you'll need twice as muchPrinciples of Chemistry II © Vanden BoutBack to BufferspKa = pH - log[A-][HA]Ka =[H+][A-][HA]This is the same equation!Let's look at the second onePrinciples of Chemistry II © Vanden BoutKa =[H+][A-][HA]If [HA] = [A-], then [H+] = Kaor we could look at it asif [H+] = Ka, then [HA] = [A-]if [H+] > Ka, then [HA] > [A-] "too many" protonsif [H+] < Ka, then [HA] < [A-] "too few" protonsPrinciples of Chemistry II © Vanden BoutWhy should I careProteins have lots of acid and base groupsNH2COHOBlood pH = 7.5NH3+CO-OpKa =4pKa =12Principles of Chemistry II © Vanden BoutWe want to "Buffer" against pH changedemoAdd NaOH to water and the pH shoots up to 12Add NaOH to mixture of acetic acid and sodium acetate and the pH doesn't change at allPrinciples of Chemistry II © Vanden BoutNaOH added to waterWater. Add 10-3 moles of OH- to the solutionThe [OH-] = 10-3 pOH = 3 pH =11Principles of Chemistry II © Vanden BoutNaOH added to bufferinitial concentration of [HA] = 0.1 Minitial concentration of [A-] = 0.1 Madd .001 moles of NaOH to 1L of solutionconcentration of [HA] = .1 -.001 = 0.099concentration of [A-] = .1 + .001 = .101Ka =[H+][A-][HA]=[H+](.101)0.09910-4.75pH = 4.76Principles of Chemistry II © Vanden BoutBuffer before adding NaOH pH = 4.75after adding NaOH pH = 4.76Water before adding NaOH pH = 7after adding NaOH pH = 3the only way to change the pH of the buffer system dramatically is to add enough acid or base to substantially change either the HA or A- concentrationsPrinciples of Chemistry II © Vanden BoutStrong 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 MPrinciples of Chemistry II © Vanden BoutpH changes rapidlybecause the total amount of H+ OH- is very small between pH 3 and pH
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