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General Chemistry IIAcid-Base ChemistryModule 3Strong Acids and BasesStrong acids and bases are completely dissociated in waterHCl + H2O H3O+ + Cl-2 H2O H3O+ + OH-[H+]Total = [H+]acid + [H+]waterIf [H+]acid > 10-6 M then [H+]water can be neglected since [H+]water 10-7 M Strong Acid or Base pHCalculate the pH of a 0.057 M solution of HClHCl + H2O H3O+ + Cl-CHCl = 0.057 M and is >> CH+ water at 10-7 MpH = -log[H3O+] = -log[5.7 x 10-2] = 1.24

For a strong base make the same assumptions but in terms of [OH-]Calculate pOHThen use pH + pOH = 14 to find pHWeak Acid-Base Equilibria The auto-ionization of H2O in equilibria involving strong acids and bases is negligible and can be ignored, making calculations much simpler.

However, if [H3O+] < 10-6 M, ignoring the auto-ionization of H2O will lead to appreciable errors. In these cases, it must be explicitly considered.

HA(aq) + H2O() H3O+(aq) + A-(aq)Ka = [H3O+][A-] / [HA]2 H2O() H3O+(aq) + OH-(aq)Kw = [H3O+][OH-] = 1.0 x 10-14 at 25CWeak Acids or BasesWeak acids and bases are only partially dissociated in solution and Cacid/base CH+/OH-Calculate pH of a 0.1 M solution of acrylic acid for which Ka = 5.56 x 10-5HAcr + H2O H3O+ + Acr-Ka = [H3O+][Acr-] / [HAcr][H3O+] = Ka[HAcr] / [Acr-] = ?

Weak Acid pH Using ApproximationsHAcr + H2O H3O+ + Acr-(1) From stoichiometry [H3O+] = [Acr-](2) For a weak acid CHAcr = [HAcr]originalKa = [H3O+][Acr-] / [HAcr]Ka = [H3O+]2 / [HAcr]original[H3O+] = (5.56 x 10-5 x 0.1) pH = 2.63

Test CHAcr = [HAcr]original [H3O+] = 0.10 0.0023CHAcr = 0.0976 (corresponding relative error 2.3%)Relative error 5% considered acceptable

Weak Acid pH without ApproximationsCalculate pH of a 0.1 M solution of acrylic acid for which Ka = 5.56 x 10-5HAcr + H2O H3O+ + Acr-t = 0[HAcr] = 0.1 Mt = [HAcr] = 0.1 x [H3O+] = [Acr-] = xKa = [H3O+][Acr-] / [HAcr] = x2 (0.1 x)x2 + 5.56 x 10-5x - 5.56 x 10-6 = 0x = 2.39 x 10-3 = [H3O+] pH = 2.62Weak Base Given pKaAcidBaseHA H+ + A-A- + H2O HA + OH-Ka = [H+][A- ] / [HA]Kb = [HA][OH-] / [A-]

Ka x Kb = ([H+][A- ] / [HA]) x ([HA][OH- ] / [A-])Ka x Kb = [H+][OH-] = KWpKa + pKb = pKW = 14

Given a value for pKa you can always find the value for pKb using the relationship pKa + pKb = 14Fractional DissociationThe fraction of a weak acid present in the form A-HA + H2O H3O+ + A-= [A-] / ([A-] + [HA])= [A-] / CHA

Calculate the % ionization of hydrocyanic acid with Ka = 4 x 10-10 in a 0.15 M solution. HCN(aq) + H2O() H3O+(aq) + CN-(aq)at equilibrium 0.15 M 7.8 x 10-6 M 7.8 x 10-6 M% ionization = (7.8 x 10-6) 100%/ (0.15)= 0.0052% 10Example: In 0.12 M solution, a weak acid HY, is 5.0% ionized. Calculate the ionization constant for the weak acid. HY(aq) + H2O() H3O+(aq) + Y-(aq)Ka = [H3O+][Y-] / [HY]HYH3O+Y-t=00.1200change0.12(0.95)0.12(0.05)0.12(0.05)at equilibrium0.110.00600.0060Ka = (0.0060)(0.0060) / (0.114) = 3.2 x 10-4 11Example: The pH of a 0.10 M solution of a weak acid, HA, is found to be 2.97. What is the value of its ionization constant? HA(aq) + H2O() H3O+(aq) + A-(aq)Ka = [H3O+][A-] / [HA][H3O+] = 10-2.97 = 1.07 x 10-3MHAH3O+A-t=0 0.1000change0.10-xxxat equilibrium0.09891.07 x 10-31.07 x 10-3

Ka = (1.07 x 10-3)(1.07 x 10-3) / (0.0989) = 1.2 x 10-5 Solution of a Salt of a Weak Acid or Base A solution of a salt of a weak acid or base is not neutral12

.Consider the equilibrium for the salt trimethylammonium chloride which results in a slightly acidic solution as a result of the following reactions

(CH3)3NHCl(aq) + H2O() (CH3)3NH+(aq) + Cl-(aq)(CH3)3NH+(aq) + H2O() H3O+(aq) + (CH3)3N(aq)Solution of a Salt of a Weak Acid or Base13Dissolving NaF in water results in a slightly basicsolution as a result of the following reactions.

NaF(aq) + H2O() Na+(aq) + F-(aq)F-(aq) + H2O() HF(aq) + OH-(aq)

Dissolving CH3CO2Na in water results in a slightlybasic solution as a result of the following reactions.

CH3CO2Na(aq) + H2O() Na+(aq) + CH3CO2-(aq)CH3CO2-(aq) + H2O() CH3CO2H(aq) + OH-(aq)14Salts of a strong acid or base Dissolving NaCl in water has no effect on the pH ofthe solution

NaCl(aq) + H2O() Na+(aq) + Cl-(aq)Cl-(aq) + H2O() no appreciable reaction

HCl is a strong acid, so Cl- is ineffective as a base,and the NaCl solution remains neutral. pH of a salt of a weak acid or baseWhat is the pH of a solution containing 0.10 M sodium phenate (Ka = 1.4 x 10-10 for phenol)NaOPh Na+ + OPh-OPh- + H2O HOPh + OH-pKa + pKb = 14 and pKa = 9.854 so Kb = 7.13 x 10-5Kb = [HOPh][OH-] / [OPh-][HOPh] = [OH-] from stoichiometryAssuming [OPh-] [NaOPh] = 0.10 M7.13 x 10-5 = [OH-]2 / 0.1 (by substitution into Kb expression) [OH-] = 2.67 x 10-3 (assumption test[HOPh] = [OH-] > [OH-], so that we can neglect [OH-]

Then equation (2) becomes[H3O+] [A-](2')

We also know that acetic acid is a weak acid (Ka = 1.8x10-5), so HA should only be slightly dissociated.

Thus, [HA] >> [A-], so that we can neglect [A-]The equation (1) becomes[HA] 0.010(1')

Approximations Rearranging the expression for Ka gives

[H3O+][A-] = Ka[HA] = 1.8 x 10-5[HA]

Substituting in for the concentration of [A-] gives

[H3O+][H3O+] = (1.8 x 10-5)(0.010 ) = 1.8 x 10-7

Taking the square root of this expression gives[H3O+] = 4.2 x 10-4 M[A-] = 4.2 x 10-4 M[OH-] = Kw/[H3O+] = 2.4 x 10-11 MApproximations Using the mass balance equation and the concentrationdetermined for A- gives

[HA] = (0.010 4.2 x 10-4)M = 0.0096 M

Check assumptions[H3O+] >> [OH-]yes[HA] >> [A-]4.2% error, okay

If the resulting concentrations violate assumptions, thenyou must resolve the problem without the assumption(s)that is (are) not satisfied. 27General Approach to Approximation Solutions Make as many approximations as you can on the mass and charge balances by neglecting any concentrations (expected to be small) when it is added to another (large one).Solve the simpler approximate system of equations.Check whether your results are consistent with your approximations.If an approximation fails to provide an answer with the desired precision, leave the equation in its exact form and solve the more complicated set of equations.Check your results. An approximation is acceptable if it introduces an error of < 5%.

28Weak Base Equilibria The treatment of weak bases is the same as that for weak acids.Aqueous ammonia and its organic derivatives, the amines, are the only common soluble weak bases.

Calculations for basic solutions are treated the same way as acidic solutions.29Example: Calculate the concentrations of the various species in 0.15M ammonia, NH3, solution. NH3(aq) + H2O() NH4+(aq) + OH-(aq)Kb = [NH4+][OH-] / [NH3] = 1.8 x 10-5NH3NH4+OH-t=00.1500change0.15-xxx

Kb = x2 / 0.15-x x2 / 0.15 = 1.8 x 10-5x = {1.8 x 10-5(.15)}1/2 = 1.64 x 10-3 M30Example: Calculate the concentrations of the various species in 0.15 M ammonia, NH3, solution. NH3NH4+OH-at equilibrium0.15 M1.64 x 10-3 M1.64 x 10-3 M

[H3O+] = Kw/[OH-] = (1.0 x 10-14)/(1.64 x 10-3) = 6.25 x 10-12 M

Check assumptions (i.e., ignoring autoionization of H2O and that ionization of NH3 is very minor)Kb = (1.64 x 10-3)2 / (0.15)= 1.8 x 10-5

The ionization constant for aqueous ammonia is the equal to the constant for acetic acid. This tells us that NH3 and CH3CO2H ionize to the same extent in aqueous solutions of the same concentration.31Example: The pH of an aqueous ammonia solution is 11.37. Calculate the molarity of the aqueous ammonia solution. pOH = 14.00 pH = 2.63 = -log10[OH-][OH-] = 10-2.63 = 2.34 x 10-3 MNH3(aq) + H2O() NH4+(aq) + OH-(aq)Kb = [NH4+][OH-] / [NH3] = 1.8 x 10-5NH3 NH4+ OH-x-y y y

at equilibrium x-2.34 x 10-3 M 2.34 x 10-3 M 2.34 x 10-3 MKb = (2.34 x 10-3)2 / (x - 2.34 x 10-3) = 1.8 x 10-5x = {(2.34 x 10-3)2 / 1.8 x 10-5} + 2.34 x 10-3 = 0.31M32Buffers A buffer solution is any solution that maintains an approximately constant pH despite small additions of acid or base and maintains a constant pH when diluted.The most common kinds of buffer solutions are. A solution composed of a weak acid and a soluble salt of that acid.A solution composed of a weak base and a soluble salt of that base.A solution of a salt of a polyprotic acid33Buffers Buffer solutions play important roles in controlling the solubility of ions in solution and in maintaining the pH in biochemical and physiological processes.Human blood has a pH of approximately 7.4. This pH is maintained by carbonate, phosphate and protein buffer systems.

Careful control of the pH of our blood is very important, if the pH drops below 7.0 or rises above 7.8 the result is rapid death. 34Buffers HA(aq) + H2O() H3O+(aq) + A-(aq)Ka = [H3O+][A-] / [HA]or[H3O+] = Ka[HA] / [A-]pH = pKa + log [A-]) / [HA])The key to effective buffer action is to keep the [HA] and [A-] nearly equal and fairly large.How does a buffer resist change in pH?The strong acid or base added to a buffer solution is consumed by [A- ] or [HA]. As long as [A-] or [HA] is not fully consumed the log term [A-] / [HA] does not change very much and therefore pH does not change very much 35Henderson-Hasselbach Equation The Henderson-Hasselbach equation for solutions of a weak acid and a salt of that weak acid (or solutions of a base and a salt of that weak base)pH pKa + log10[A-]/[HA]pH pKa + log10[B]/[BH+]

pKa in the 2nd equation refers to the pKa of BH+Henderson-Hasselbach Equation pH pKa + log10[A-]/[HA]The pH at which a buffer operates depends on the ratio of conjugate acid and base in the buffer solution and on the ionization constant of the weak acid or base involved

For 1% [A-] or [HA] in a buffer (ratio 1:100) this corresponds to a range of pH pKa 2

Effective buffering capacity is usually restricted to pKa 1 corresponding to a ratio of 10:1 or 1:10 for [A-] and [HA] 37Henderson-Hasselbach Equation The correct Henderson-Hasselbach equation includes activity coefficients.

Because K depends on T and activity rather than concentration, ignoring these effects will sometimes lead to significant errors.Composition of a BufferWhat ratio of sodium acetate and acetic acid (Ka = 1.8 x 10-5 must be taken to give a buffer solution of pH =5NaOAc Na+ + OAc-HOAc + H2O [H3O+] + [OAc-] Ka = [H3O+] [OAc-] / [HOAc] [OAc-] / [HOAc] = Ka / [H3O+] = 1.8 x 10-5 / 1 x 10-5[OAc-] / [HOAc] = 1.8 39Example: Calculate the concentration of H3O+ and the pH of a solution that is 0.15 M in acetic acid and 0.15 M in sodium acetate. pH pKa + log10([A-]0/[HA]0) = -log10(1.8 x 10-5) +log10(0.15/0.15) = 4.7[H3O+] = 1.8 x 10-5 M

Compare this to the solution with only 0.15 M acetic acid, where we found that [H3O+] = 1.6 x 10-3 M, almost 100 times as large as in the buffer solution. ???40If 0.020 mole of gaseous HCl is added to1.00 L of solution that is 0.200 M in aqueous ammonia and 0.100 M in ammonium chloride, how much does the pH change? Assume no volume change due to the addition of gaseous HCl.pH of original buffer solutionpH pKa + log10([NH3]0/[NH4+]0) Kb of NH3=1.7610-5 = -log10(5.68 x 10-10) + log10(0.200/0.100) = 9.546After HCl is addedHCl(aq) + NH3 (aq) NH4+ (aq) + Cl- (aq)[NH3] = (0.200 - 0.020) M = 0.180 M[NH4+] = (0.100 + 0.020) M = 0.120 MpH after the HCl is addedpH pKa + log10[NH3]0/[NH4+]0 = -log10(5.68 x 10-10) + log10(0.180/0.120) = 9.421DpH = 9.421-9.546= -0.125 (correct significant figures would be two digits)41Preparation of Buffer Solutions When designing a buffer solution, chose an acid so that the desired pH pKa and then adjust the initial [HA]0 and [A-]0 to achieve the desired pH.

pKa values for commonly used buffers are given in Table 9-2.

The absolute concentrations of [HA]0 and [A-]0 are not important in pH determination, just their ratio.42Preparation of Buffer Solutions However, absolute concentrations affect the ability of the buffer solution to resist pH changes when acid or base is added.

The greater the initial concentrations of [HA]0 and [A-]0, the smaller the change in pH induced by addition of acid or base.43Buffer Capacity The buffer capacity, , is a measure of how well a solution resists changes in pH when strong acid or base is added. = dCb/dpH = -dCa/dpHwhere Ca and Cb are the # of moles of strong acid and strong base per liter needed to produce a unit change in pH.

*The larger is the greater the buffer capacity and the better the buffer.44Buffer Capacity The buffer capacity said to be exhausted if enough strong acid (base) is added to use up the original amount of the weak base (acid).Buffer solutions are often prepared by mixing other solutions.

When solutions are mixed, the concentrations of the dissolved species change because the volume changes.45Example: Calculate the concentration of H3O+ and the pH of the solution prepared by mixing 200 mL of 0.150 M acetic acid and 100 mL of 0.100 M sodium hydroxide solutions.

NaOH(aq) in H2O () Na+(aq) + OH- (aq)OH-(aq) + CH3CO2H (aq) CH3CO2- (aq) + H2O()# moles = Molarity x Volumet=0 (mol) 0.01000.03000mixing(M) 0.03330.100change 0.0333-x0.100-xxafter reaction 00.06670.0333pH pKa + log10[CH3CO2-]0/[CH3CO2H]0 = -log10(1.8 x 10-5) + log10(0.0333/0.0667) = 4.44Preparation of a Buffer from SolutionsCalculate the pH of a buffer solution prepared by mixing 200 mL of 0.150 M acetic acid and 100 mL of 0.100 M sodium hydroxide solutions. (Ka HOAc = 1.8 x 10-5) On mixing NaOH and HOAc react to form NaOAc NaOH + HOAc NaOAc + H2OThere are fewer moles of NaOH than HOAC and the solution will contain only HOAc and NaOAc after mixingMoles NaOAc = Moles NaOH (= Molarity x Volume) = 0.1 x 0.1 = 0.010Moles HOAc = HOAcinitial NaOAc(moles NaOAc = moles of HOAc reacted)HOAc = (0.2 x 0.15) 0.010 = 0.020When the solutions are mixed the total volume = 200 + 100 = 300 mL[NaOAc] = 0.010 / 3 = 0.00333 and [HOAc] = 0.00667For a buffer solutionpH = pKa + log[OAc] / [HOAc]pH = -log (1.8 x 10-5) + log (0.00333 / 0.00667) = 4.443pH = 4.44 (correct number of significant figures)

47Example: Calculate the number of moles of solid ammonium chloride, NH4Cl, that must be used to prepare 1.00 L of a buffer solution that is 0.10 M in aqueous ammonia, and that has a pH of 9.15.

NH3(aq) + H2O() NH4+ (aq) + OH- (aq)Ka(NH4+) = 5.68 x 10-10or pKa = 9.245

pH pKa + log10([NH3]/[NH4+])9.15 = 9.245 + log10(0.10/[NH4+])log10(0.10/[NH4+]0) = -0.095[NH4+] = 0.10 M / (10-0.095) = 0.12 MSo moles of ammonium chloride = [NH3] + [NH4+] = 0.22 M