4 Fig 16.4 Relative strengths of some conjugate acid-base pairs The strongerthe acid…the weaker itsconjugate base.
5 The Ion Product of Water Kc =[H+][OH-][H2O]H2O (l) H+ (aq) + OH- (aq)[H2O] = 55.6 M= constantKc[H2O] = Kw = [H+][OH-]Ion-product constant (Kw) - the product of the molar concentrations of H+ and OH- ions at a particular temperature.Solution is:[H+] = [OH-]neutralAt 25°C:Kw = [H+][OH-] = 1.0 x 10-14[H+] > [OH-]acidic[H+] < [OH-]basic
6 pH – A Measure of Acidity pH = −log [H+]pH = −log [H3O+]Solution is:At 25°Cneutral[H+] = [OH-][H+] = 1 x 10-7pH = 7acidic[H+] > [OH-][H+] > 1 x 10-7pH < 7basic[H+] < [OH-][H+] < 1 x 10-7pH > 7pH[H+]
7 Fig 16.5 H+ concentrations and pH of common substances pOH = -log [OH-][H+][OH-] = Kw = 1.0 x 10-14−log [H+] – log [OH-] = 14.00pH + pOH = 14.00
8 The pH of rainwater collected in a certain region of the northeastern United States on a particular day was What is the H+ ion concentration of the rainwater?pH = -log [H+][H+] = 10-pH== 1.5 x 10-5 MThe OH- ion concentration of a blood sample is 2.5 x 10-7 M. What is the pH of the blood?pH + pOH = 14.00pOH = -log [OH-]= -log (2.5 x 10-7)= 6.60pH = – pOH = – 6.60 = 7.40
9 Other “p” FunctionsThe “p” in pH tells us to take the negative base - 10 logarithm of the quantity (in this case, hydronium ions)Some similar examples:pOH = -log [OH−]pKw = -log KwpCl = -log [Cl−]
10 How Do We Measure pH? For less accurate measurements: Litmus paper “Red” paper turns blue above ~pH = 8“Blue” paper turns red below ~pH = 5Indicator:
11 How Do We Measure pH? Fig 16.6 Digital pH meter For more accurate measurements:pH meter, which measures the voltage in the solutionFig Digital pH meter
15 Table 16.2 Some Weak Acids in Water at 25 °C 16.5
16 Calculating Ka from the pH Sample Exercise p 682The pH of a 0.10 M solution of formic acid, HCOOH, at 25Cis Calculate Ka for formic acid at this temperature.HCOOH (aq) H+ (aq) + HCOO- (aq)Ka =[H+][HCOO-][HCOOH]pH = -log [H+]2.38 = -log [H+]-2.38 = log [H+]= 10log [H+] = [H+]4.2 10-3 = [H+] = [HCOO-]
18 For a monoprotic acid HA: Concentration ionizedOriginal concentrationx 100%Percent ionization =For a monoprotic acid HA:Percent ionization =[H+][HA]0x 100%[HA]0 = initial concentrationFig 16.9The more dilute the acid,the greater the percentionization:
19 What is the pH of a 0.50 M HF solution (at 25°C)? Ka =[H+][F-][HF]= 6.8 x 10-4HF (aq) H+ (aq) + F- (aq)HF (aq) H+ (aq) + F- (aq)Initial (M)0.500.000.00Change (M)-x+x+xEquilibrium (M)xxxKa =x2x= 6.8 x 10-4If [HF] > 100 Ka0.50 – x 0.50Ka x20.50= 6.8 x 10-4x2 = 3.40 x 10-4x = M[H+] = [F-] = MpH = -log [H+] = 1.72[HF] = 0.50 – x = 0.48 M
20 When can I use the approximation? If [HF] > 100 KaThen 0.50 – x 0.50Let’s determine error introduced:0.018 M0.50 Mx 100% = 3.6%Less than 5%Approximation ok.x = 0.018What is the pH of a 0.05 M HF solution (at 25°C)?Ka x20.05= 6.8 x 10-4x = M0.006 M0.05 Mx 100% = 12%More than 5%Approximation not ok.Must solve for x exactly using quadratic equation...
21 Solving weak acid ionization problems: Identify the major species that can affect the pH.In most cases, the autoionization of water can be ignored.Ignore [OH¯] because it is determined by [H+].Use ICE table to express the equilibrium concentrations in terms of single unknown x.Write Ka in terms of equilibrium concentrations. Solve for x by the approximation method. If approximation is not valid, solve for x exactly.Calculate concentrations of all species and/or pH of the solution.
22 What is the pH of a 0.122 M monoprotic acid whose Ka is 5.7 x 10-3?HA (aq) H+ (aq) + A- (aq)Initial (M)0.1220.000.00Change (M)-x+x+xEquilibrium (M)xxxKa =x2x= 5.7 x 10-3Is [HF] > 100 Ka ?NO!!Approximation not ok.
23 Ka =x2x= 5.7 x 10-3x X 10-3 x – 6.95 X 10-4 = 0-b ± b2 – 4ac2ax =ax2 + bx + c =0x =x =HA (aq) H+ (aq) + A- (aq)Initial (M)Change (M)Equilibrium (M)0.1220.00-x+xxx[H+] = x = MpH = -log[H+] = 1.625
24 Polyprotic Acids Have more than one ionizable proton If difference between the Ka1 and subsequent Ka values > 103, the pH generally depends only on the first dissociation.