Chapter 8 Monoprotic Acid-Base Eqlilibria

8-1 Strong Acids and Bases Completely dissociates : [H3O+] or [OH-] equals concentration of strong acid or base (8-1) Ex. What is the pH of a 0.1M solution of HCl? Ex. What is the pH of a 0.1M solution of KOH? Ka of strong acids Acid Ka HCl 103.9 HBr 105.8 HI 1010.4 HNO3 101.4 [OH-], M [H+], M pH 10-3.00 10-11.00 11.00 10-4.00 10-10.00 10.00 10-5.00 10-9.00 9.00

The Dilemma What is the pH of 1.0x10-8 M KOH? How can a base produce an acidic solution? ⇒ Wrong Assumption!! For large concentration of strong acid or base, [H+] = [acid] or [OH-] = [base] For small concentration, must account for water dissociation ⇒ In pure water [OH-] = 1.0×10-7M, which is greater than [KOH] = 1×10-8M ⇒ Must Use Systematic Treatment of Equilibrium

The Cure Step 1: Pertinent reactions: (Completely dissociates, not pertinent) Step 2: Charge Balance: Step 3: Mass Balance: (All K+ comes from KOH) Step 4: Equilibrium constant expression (one for each reaction): Step 5: Count equations and unknowns: Three equations: Kw, MB, CB Three unknowns: K+, H+, OH- Step 6: Solve (Seeking pH, [H+]): From CB, MB: For Kw: Use quadratic equation (pH slightly basic, consistent with low [KOH])

- Three Regions depending on acid/base concentrations pHs of Strong Acids and Bases for different concentrations of Strong Acids and Bases - Three Regions depending on acid/base concentrations Low concentrations, (10-6~10-8M), H2O ionization ≈ H+,OH- of acid/base  systematic equilibrium calculation necessary High concentrations (≥10-6M), pH considered just from the added H+,OH- Very low concentrations (≤10-8M), pH=7  not enough H+, OH- added to change pH

Water Almost Never Produces 10-7 M H+ and OH- - pH=7 only true for pure water ([H+]=[OH-]=1.0×10-7 M) - Any acid or base suppresses water ionization : Follows Le Châtelier’s principal Ex. In 10-4 M HBr :In 10-4 M HBr solution, water dissociation produces only 10-10 M OH- and H+

8-2 Weak Acids and Bases Ka · Kb = Kw Weak acid/base do not completely dissociate - Dissociation constant (Ka ) for the weak acid HA: Ka (8-3) - Base Hydrolysis constant (Kb ) for the weak base B: Kb (8-4) - As Ka or Kb increase  pKa or pKb decrease - Smaller pKa  stronger acid Formic acid stronger acid than benzoic acid (HA) (A-) (HA) (A-) Conjugate acid-base pair Ka · Kb = Kw (8-5)

Weak Is Conjugate to Weak The conjugate base of a weak acid is a weak base. The conjugate acid of a weak base is a weak acid. The conjugate base of a strong acid is so weak acid(Not a base in water). KaㆍKb = Kw (for conjugate acid-base pair) Ka=10-4 ⇒ Kb ? Ka=10-5 ⇒ Kb ? Using Appendix G pKa = -logKa pKa Group Ka 1.4 POH 0.04 3.51 OH 3.1×10-4 6.04 9.1×10-7 8.25 NH 5.6×10-9

A Typical Weak-Acid Problem 8-3 Weak Acid Equilibria A Typical Weak-Acid Problem General Systematic Treatment of Equilibrium - Unlike concentrated strong acid, need to account for water ionization - Find pH for a solution of a general weak acid (HA) Step 1: Pertinent reactions: Step 2: Charge Balance: (8-6) Step 3: Mass Balance: (8-7) Step 4: Equilibrium constant expression (one for each reaction): (8-8) Step 5: Count equations and unknowns: Four Equations: (8-6), (8-7), (8-8), Kw Four Unknowns:

Step 6: Solve (Not easy to solve  cubic equation results!): - Again, need to make assumptions to simplify equations - The goal is to determine [H+], so we can measure pH Make Some Initial Assumptions: - For a typical weak acid, [H+] from HA will be much greater than [H+] from H2O - If dissociation of HA is much greater than H2O, [H+] ≫ [OH-] [OH-] neglect ⇒ for CB; (8-9) From (8-9): Set [H+]=x  [A-]=x Ex) F=0.050M, Ka=1.07x10-3, pH=? For MB: Use quadratic equation For (8-8): Use quadratic equation

Step 7: Verify Approximation: Was the approximation, [H+] >>[OH-] justified ([H+] ≈ [A-])? Setting F = 0.050 M and Ka = 1.07x10-3 for o-hydroxybenzoic acid: [H+] >> [OH-] 6.8x10-3M >> 1.5x10-12M assumption is justified! Determine [OH-] from water dissociation:

Summery for Weak-Acid Problem HA A- + H+ Initial(M) F 0 0 F: Total concentration Change(M) -x +x +x Approximation: [A-]≈[H+]=x Equili.(M) (F-x) (x) (x) Assump. Weak acid: F ≫ [H+]  F - [H+] ≈ F Use quadratic equation Caution for assumption (F≫[H+]) : F/Ka > 102 → Can be assump. : Confirm when large Ka or low F : Should be justified

Ex) 0.1M CH3COOH(Ka=1.8×10-5), pH=? Ex) 0.0500 M benzoic acid, pH=? COOH Benzoic acid Ka = 6.28 ×10–5 Using approximation: F-x ≒ F

Fraction of Dissociation (8-10) Stronger acid Ex: What is the percent fraction dissociation for F = 0.050 M o-hydroxybenzoic acid (Ka = 1.07x10-3 ) ? ① The α of the stronger acid is more higher than weaker acid. ② The α of weak acid(weak electrolyte) increases as diluted. Weaker acid

Essence of a Weak-Acid Problem F: Total concentration (8-11) Approximation: [A-]≈[H+]=x Example A Weak-Acid Problem Find the pH of 0.050M trimethylammonium chloride. Solution From App. G: Ka=10-pKa = 10-9.797 = 1.59x10-10 (CH3)3NH+Cl- → (CH3)3NH+ + Cl- Ka (CH3)3NH+ = (CH3)3N + H+ (8-12) Using approximation: F-x ≒ F

8-4 Weak Base Equilibria Treatment of Weak Base (B) is Very Similar to Weak Acid - Assume all OH- comes from base and not dissociation of water B + H2O BH+ + OH- Initial(M) F 0 0 Change(M) -x +x +x Equili.(M) (F-x) (x) (x) Approx. Weak base: F ≫ [OH-]  F - [OH-] ≈ F (8-13) Use quadratic equation Caution for approx. (F≫[OH-]) : F/Kb > 102 → Can be approx. : Confirm when large Kb or low F : Should be justified

• Fraction of Dissociation(α) Ex) What is the pH of cocaine dissolved in water? (F = 0.0372 M and Kb=2.6x10-6 for cocaine) Initial(M) 0.0372 0 0 Change(M) -x +x +x Eqlili.(M) (0.0372-x) (x) (x) • Fraction of Dissociation(α) (8-14)

8-5 Buffers A buffered solution resists changes in pH when acids or bases are added - Buffer is a mixture of a weak acid(HA) and its conjugate base(A-) : Must be comparable amounts of acid & base (>0.1M) Composition of buffer solution ① Weak acid(HA) + Conjugate base (A-) ② Weak base(B) + Conjugate acid (BH+) pH dependence of the rate of cleavage of an amide bond by the enzyme chymotrypsin.

Mixing a Weak Acid and Its Conjugated Base ① HA = H++ A- ② A-+ H2O = HA + OH- HA + A- 1) Weak acid(HA) reacts “incompletely (very little)“ with a weak base(A-). 2) Very little change of moles (or concentrations) for the mixing HA and A- Consider FHA=0.10M HA (pKa=4.00) α Low α little dissociation of HA and A- Consider FA-=0.10M A- (pKb=10.00) Mixing A-  Reaction ① goes to the left Le Chatelier’s principle Mixing HA  Reaction ② goes to the left ⇒ From the approximation 1) and 2), the equilibrium concentration of HA and A- remain unchanged. [HA] ≒ FHA [A-] ≒ FA-

Henderson-Hasselbalch Equation pH of buffer solution(HA + A-) Take log of both sides: Rearrange: pH pKa (8-16) pH of buffer solution(B + BH+) (8-17) When activities are includes, (8-18)

Properties of the Henderson-Hasselbalch Equation Determines pH of buffered solution ⇒ Need to know ratio of conjugate [acid, HA] and [base, A-] If [A-] = [HA]  pH = pKa All equilibria must be satisfied simultaneously in any solution at equilibrium Only one concentration of H+ in a solution [A-]/[HA] pH 100:1 pKa + 2 10:1 pKa + 1 1:1 pKa 1:10 pKa - 1 1:100 pKa - 2

Buffer in Action Ex. Tris buffer BH+ B (This form is “tris”) Tris (hydroxymethyl)aminomethane BH+ (pKa=8.072) B (This form is “tris”) - Dissociation of Tris hydrochloride(BH+Cl– ) BH+Cl–  BH+ + Cl– BH+ B + H+

- The pH of a buffer is nearly independent of volume Example A Buffer Solution Find the pH of 1.00-L aqueous solution containing 12.43g of tris(FM 121.135, B) plus 4.67 g of tris hydrochloride(FM 157.596, BH+) (pKa = 8.07). Solution [B]=(12.43g/121.135g/mol)/1.00L = 0.1026 M [BH+]=(4.67g/157.596g/mol)/1.00L = 0.0296 M - The pH of a buffer is nearly independent of volume Example Effect of Adding Acid to a Buffer If we add 12.0 mL of 1.00 M HCl to the solution in previous solution, what will be the new pH? pH does not change very much (ΔpH=-0.20) Solution B + H+ → BH+ Initial(mol) 0.1026 0.0120 0.0296 Final(mol) 0.0906 0 0.0416 (0.1026-0.0120) (0.0296+0.0120)

Calculating How to Prepare a Buffer Solution Example Calculating How to Prepare a Buffer Solution How many milliters of 0.500 M NaOH should be added to 10.0 g of trisH+(BH+) to give a pH of 7.60 in a final volume of 250 mL? Solution Adding OH-  reduce [BH+] and increase [B] Moles of trisH+(BH+) = (10.0 g)/157.596 g/mol) = 0.0635 mol BH+ + OH- → B + H2O Initial(mol) 0.0635 x 0 Final(mol) 0.0635-x 0 x Mole of OH- = MV ⇒ 0.0160 mol=(0.500M)(a mL) ∴ a=mL of NaOH = 32.0 mL ⇒ dilute to 250.0 mL

Preparing a Buffer in Real Life! 0.1 M tris containing Tris buffer (pH = 7.60, 1.0 L) Prepare 0.1 mol trisH+ using beaker (~800mL) - Monitor pH using pH meter - Add NaOH solution until the pH is exactly 7.60 Transfer to volumetric flask of 1.0 L Dilute to the mark (1.0L) with water and mix. The pH should be adjusted by pH meter!!! (Do not simply add the calculated quantity of NaOH)

Buffer Capacity Why Does a Buffer Resist Changes in pH? - Strong acid or base is consumed by B or BH+ - Buffer Capacity (β): measure of a solutions resistance to pH change(“+” value) (8-19) :where Ca and Cb are the number of moles of strong acid and strong base per liter needed to produce a unit change in pH - Greater β  more resistant to pH change - Maximum capacity to resist pH change occurs at pH=pKa Choosing a Buffer - Choose a buffer with pKa as close as possible to desired pH - Useful buffer range is pKa ± 1 pH units - Buffer pH depends on temperature and ionic strength  activity coefficients

[HA]=[A-]→[A-]/[HA]=1 ⇒ pH = pKa : max β Buffer capacity depends on: - Absolute concentration of HA and A- - Relative concentration(molar ratio) of HA and A- [HA]=[A-]→[A-]/[HA]=1 ⇒ pH = pKa : max β [A-]/[HA] pH 100:1 pKa + 2 10:1 pKa + 1 1:1 pKa 1:10 pKa - 1 1:100 pKa - 2 Maximum buffer capacity : pH = pKa Effective buffer range : pKa ± 1

Buffer pH Depends on Ionic Strength and Temperature - Changing ionic strength (μ) changes pH. (∵ Buffer solution: high concentration (ionic strength ≠ 0, <1)  reduce effective concentration  use activity (A)) - Changing temperature changes pH. (∵ pKa depends on temperature.) Tris buffer: pH 8.07 (25 0C) pH 8.7 (4 0C) pH 7.7 (37 0C) When What You Mix Is Not What You Got : In dilute solution or extremes of pH, don’t use Henderson-Hasselbalch Eq. In dilute solution (∵increase α values of HA and A-) In extremes of pH, [H+]≫[OH-] or [H+]≪[OH-] FHA ≠ [HA] FA- ≠ [A-]

A Dilute Buffer Prepared from a Moderately Strong Acid Example A Dilute Buffer Prepared from a Moderately Strong Acid Mixing of 0.0100 mol HA (pKa=2.00) and 0.0100 mol of A- in 1.00 L  pH = ? Solution Initial(M) 0.010 0 0.010 Change(M) -x +x +x Eqlili.(M) (0.010-x) (x) (0.010+x) (8-22) Use quadratic equation (correct) Applying approximation ([HA]≒FHA and [A-]≒FA-) : Henderson-Hasselbalch Eq. (incorrect) HA is too strong and the concentrations are too low for HA and A- ⇒ [HA]≠FHA and [A-] ≠FA-

Effect of dilution of a buffer solution # Very diluted buffer solution changes pH. Dilution pH constant Constant Dilution pH increase dilution - Dilution will not alter the pH of a buffer solution. (It will, however, decrease the buffer capacity, since the concentration has been reduced.)