The Henderson-Hasselbalch Equation. The Henderson-Hasselbalch Equation the significance of pH the predominant solution species The significance of pH in solutions which contain many different acid/base pairs Diprotic Acids
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The Henderson-Hasselbalch Equation The Henderson-Hasselbalch Equation the significance of pH the predominant solution species The significance of pH in solutions which contain many different acid/base pairs Diprotic Acids Triprotic Acids Example - calculating concentrations of all phosphate containing species in a solution of KH2PO4 at known pH pH buffer solutions - choosing the conjugate acid/base pair - calculating the pH
[A-] log10 Ka = log10 [H+] + log10 [HA] [A-] - pKa = - pH + log10 [HA] [A-] pH = pKa + log10 [HA] Henderson-Hasselbalch Equation [H+] [A-] HA = H+ + A- Ka = [HA] An especially convenient form of the equilibrium equation is obtained by re-writing the equilibrium expression using logs - Henderson-Hasselbalch Equation
[A-] pH = pKa + log10 [HA] Use of the Henderson-Hasselbalch Equation In most practical cases, the pH of the solution is known either from (1) direct measurement with a pH meter (2) use of a pH buffer in the solution When the pH is known, the H.-H. Equation is much more convenient to use than the equilibrium constant expression. It immediately gives the ratio of concentrations of all conjugate acid/base pairs in the solution. Henderson-Hasselbalch Equation calculate ratio of (base/acid) concentrations measured or set by buffer known (from tabulations)
= =1 [A-] [base] [A-] [HA] [acid] [HA] pKa and the Dissociation of Weak Acids For any conjugate acid/base pair, HA = H+ + A- pH = pKa + log10 Note that pH = pKa when because log10 (1) = 0
mostly CH3COOH Mostly CH3COO- pKa=4.75 CH3COOH = H+ + CH3COO- pKa = 4.75 0 7 14 When pH < pKa, acetic acid is mostly protonated. pH When pH > pKa, acetic acid is mostly deprotonated. When pH = pKa, concentrations of the protonated and deprotonated forms are equal. The Use of pH plots
pH, pKa and % Dissociation When the pH = pKa, half the conjugate acid/base pair is in the protonated form, half is de-protonated. If the pKa = 4.7 (as for acetic acid): At pH 4.7 [CH3COO-] /[CH3COOH]= 1 (equal) pH 5.7 [CH3COO-] /[CH3COOH]= 10 (mostly de-protonated) pH 3.7 [CH3COO-] /[CH3COOH]= 0.1 (mostly protonated)
pH = pKa + log10 [CH3COO-] [CH3COOH] An Exercise in % Dissociation A 0.050 M acetic acid solution is made pH 7.00 with added NaOH. Find [CH3COOH], [CH3COO-], and [H+] in this solution. ratio = 180 7.00 4.75 Thus, [CH3COO-] ≈ 0.050 M [CH3COOH] ≈ 2.8 x 10-4 M most of the acetic acid is dissociated (% undissociated is 0.56%)
pH = pKa + log10 r = pH = pKa + log10 (r) where [base] [base] [acid] [acid] How to Use the [base] / [acid] Ratio [acetic acid] + [acetate] = 0.050 M [acetate] / [acetic acid] = r [acetic acid] = r * [acetate] [acetic acid] (1 + r) = 0.050 M [acetic acid] = 0.050 M / (1 + r) = 2.8 x 10-4 M [acetate] = 0.050 M - 2.8 x 10-4 M ≈ 0.050 M
Summary: Significance of the pH The solution may contain many conjugate acid/base pairs (biological solutions usually do). In order to reproduce a sample, you need to reproduce the pH. This guarantees that all conjugate acid/base pairs will have the same ratio of protonated/deprotonated concentrations as in the original sample. 2. When the pH is known, you can readily calculate the ratio of (protonated/deprotonated) forms of any acid for which you know the pKa.
Ratio of protonated and deprotonated species Knowledge of the pH completely determines the state of protonation / deprotonation of every Bronsted conjugate acid/base pair in solution. An example: A solution at pH 6.9 contains lactic acid (pKa = 3.9). Is lactic acid predominantly in the protonated form or the deprotonated form (lactate ion)? Ans: Predominantly deprotonated. The ratio [Lac-]/[HLac] = 103
0 7 14 pH Diprotic Weak Acid - H2CO3 H2CO3 = H+ + HCO3-pKa1 = 6.4 HCO3-= H+ + CO32- pKa2 = 10.3 pKa1=6.4 pKa2=10.3 mostly H2CO3 mostly CO32- mostly HCO3-
0 7 14 pH Triprotic Weak Acid - H3PO4 H3PO4 = H+ + H2PO4-pKa1 = 2.12 H2PO4- = H+ + HPO42-pKa2 = 7.21 HPO42- = H+ + PO43- pKa3 = 12.67 pKa1=2.12 pKa2=7.21 pKa3=12.67 mostly H3PO4 mostly H2PO4- mostly HPO42- mostly PO43-
0 7 14 pH Effect of pH on Solution Composition: H3PO4 pKa1=2.12 pKa2=7.21 pKa3=12.67 mostly H3PO4 mostly H2PO4- mostly HPO42- mostly PO43- Problem: A solution is prepared by dissolving 0.100 mole of NaH2PO4 in water to produce 1.00 L of solution. The pH is then adjusted to pH 8.50 with NaOH. What are the concentrations of H3PO4, H2PO4-, HPO42-, PO43-, and H+?
[base] [acid] Calculating the Concentrations pH = pKa + log (1) Apply H-H eqn to [H2PO4-] and [HPO42-] using pKa2 : 8.50 = 7.21 + log10 ( [HPO42-] / [H2PO4-] ) This gives [HPO42-] / [H2PO4-] = 20 (2) Apply H-H eqn to [H3PO4] and [H2PO4-] using pKa1 : 8.50 = 2.12 + log10 ( [H2PO4-] / [H3PO4] ) This gives [H2PO4-] / [H3PO4] = 2.4 x 106 Apply H-H eqn to [PO43-] and [HPO42-] using pKa3 : (3) 8.50 = 12.67 + log10 ( [PO42-] / [HPO42-] ) This gives [PO43-] / [HPO42-] = 6.8 x 10-5
0 7 14 pH Answers to the Exercise pKa1=2.12 pKa2=7.21 pKa3=12.67 mostly H3PO4 mostly H2PO4- mostly HPO42- mostly PO43- solution Ans: [HPO42-] = 0.95 x10-1 M [PO43-] = 1.6 x10-4 * [HPO42-] [H3PO4] = 2.0 x106 * [H2PO4-] [H2PO4-] = 0.047 x10-1 M [H3O+] = 3.2 x10-8 M
(a) Calculate the pH of a 500 mL solution prepared from: 0.050 mol of acetic acid and 0.020 mol sodium acetate. HAc = H+ + Ac- pKa= 4.75 ? [Ac-] pH = pKa + log10 [HAc] 4.75 ? ? Problem
Problem Suppose that 0.010 mol NaOH is added to the buffer of part (a). What is the pH? (b) HAc + OH- = H2O + Ac- pKa = 4.75 (0.020 + 0.010) / 0.50 [Ac-] pH = pKa + log10 [HAc] (0.050 - 0.010) / 0.50 4.75 4.63 As long as the buffer capacity is not exceeded, the change of pH is small, in this case, 4.35 to 4.63
pH Buffers 1. pH buffers resist a change in pH upon addition of small amounts of either acid or base. Buffer solutions should contain roughly equal concentrations of a conjugate acid and its conjugate base. The conjugate acid/base pair of the buffer should have a pKa that approximately equals the pH. For example: a buffer will result from mixing 0.1 M acetic acid and 0.1 M sodium acetate. Added OH- is neutralized by the conjugate acid Added H+ is neutralized by the conjugate base H+ + C2H3O2- = HC2H3O2 OH- + HC2H3O2 = H2O + C2H3O2-
pH = pKa + log10 5.00 4.75 [base] [acid] pH of Buffer Solutions Problem: Prepare a pH 5.00 buffer using sodium acetate and acetic acid ratio [base]/[acid] = 1.78 Any solution with this composition (i.e., this ratio of base / acid), will form a buffer, but higher concentrations provide higher buffering capacity. For example, one could use 0.178 M sodium acetate + 0.100 M acetic acid
Exercises with Buffers Use the data in Table 10.2 to design buffers at: pH 6.9 pH 9.3 pH 3.6 Find the weight of solid compounds you would use to produce 100 mL each buffer.