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Buffer Solutions pH of solution adding 0.10 M HCl to 100 mL water HCl added pH 7 0 mL 7.00 6 2 mL 2.71 5 5 mL 2.32 4 10 mL 2.04 3 20 mL 1.78 2 50 mL 1.48 1 0 1020304050 mL of 0.10 M HCl added Buffer Solutions ν A buffer helps a solution maintain its pH when acid or base is added ν A buffer must contain two components to work ν a weak acid that reacts with added base ν a weak base that reacts with added acid ν Buffers usually contain approximately equal amounts of a weak acid and its conjugate base 1 Buffer Solutions Solution that is 0.100 M CH3COOH (acetic acid) and 0.100 M NaCH3COO (sodium acetate) Find pH of buffer solution: ↔ - + CH3COOH(aq) + H2O CH3COO (aq) + H3O (aq) - + [CH3COOH] [CH3COO ] [H3O ] initial 0.100 0.100 ≈0 -x x x equil 0.100 – x 0.100 + x x Buffer Solutions Find pH of buffer solution: ↔ - + CH3COOH(aq) + H2O CH3COO (aq) + H3O (aq) - + + = [CH3COO ][H3O ] = (.100 x)x = -5 Ka 1.8 x 10 [CH 3COOH ] (.100 - x) assume x is negligible compared to .100 M x = 1.80 x 10-5 M pH = 4.74 2 Buffer Solutions Add 5 mL .10 M HCl Find pH of resulting solution Assume all acid added reacts with acetate ion to form acetic acid (remember that acids react with bases) ↔ - + CH3COOH(aq) + H2O CH3COO (aq) + H3O (aq) + -3 [H3O ] added = (5 mL)(.10 M)/(105 mL) = 4.76x10 M [CH3COOH] = 0.100 + 0.005 = 0.105 M - [CH3COO ] = 0.100 - 0.005 = 0.095 M Buffer Solutions Now let solution come to equilibrium - + [CH3COOH] [CH3COO ] [H3O ] initial .105 .095 ≈0 -x x x equil .105 – x .095 + x x - + + = [CH3COO ][H3O ] = (.095 x)x = -5 Ka 1.8 x 10 [CH 3COOH ] (.105 - x) x = 1.99 x 10-5 M pH = 4.71 3 Buffer Solutions pH of buffered solution adding 0.10 M HCl to 100 mL soln 7 HCl added pH 6 0 mL 4.74 5 mL 4.71 5 Buffered solution 10 mL 4.66 4 15 mL 4.58 3 25 mL 4.57 2 1 50 mL 4.45 0 1020304050 mL of 0.10 M HCl added Buffer Solutions ν Henderson-Hasselbach Equation ν Allows calculation of pH of a buffer if concentrations of conjugate acid and conjugate base are known ↔ + - HA(aq) + H2O H3O (aq) + A (aq) [H O+ ][A-] K = 3 a [HA] + K [HA] [H O ] = a 3 [A-] 4 Buffer Solutions ν Take -log of both sides + ⎧ K [HA]⎫ ⎛ [HA]⎞ −log[H O ] = -log⎨ a ⎬ = - log() K -log⎜ ⎟ 3 ⎩ [A-] ⎭ a ⎝ [A-] ⎠ ⎛ [HA]⎞ ⎛ [A-] ⎞ -log(K ) = pK −log⎜ ⎟ = log⎜ ⎟ a a ⎝ [A− ]⎠ ⎝ [HA]⎠ ⎛ [A-] ⎞ pH = pK + log⎜ ⎟ Henderson-Hasselbach a ⎝ [HA]⎠ Eqn Buffer Solutions ν Using the Henderson-Hasselbach Eqn, we can: ν Determine pH of a solution ν Determine ratio of conjugate base to conjugate acid to achieve specific pH ⎛ [A-] ⎞ pH = pK + log⎜ ⎟ a ⎝ [HA]⎠ 5 Buffer Solutions ν Let’s go back to problem of adding HCl to buffer solution: ν We can use H-H eqn. to make the calculations much easier [CH3COOH] = 0.100 + [HCl]added - [CH3COO ] = 0.100 – [HCl]added ⎛ [A-] ⎞ pH = pK + log⎜ ⎟ a ⎝ [HA]⎠ Buffer Solutions - VHCl [HCl] [CH3COOH] [CH3COO ] pH 5mL (.1)(5mL)/105mL .100+.005 .100-.005 = .00476 M =.105 M =.095 M 4.70 10mL (.1)(10)/110 .100+.009 .100-.009 = .00909 M =.109 M =.091 M 4.66 25mL (.1)(25)/125 .100+.020 .100-.020 = .0200 M =.120 M =.080 M 4.56 50mL (.1)(50)/150 .100+.033 .100-.033 = .0333 M =.133 M =.067 M 4.44 6 Buffer Solutions ν Buffer Capacity—the amount of acid or base that can be added to a buffer without the pH significantly changing ν Suppose we acid to a buffer solution: ν The acid will react with the conjugate base until it is depleted ν Past this point, the solution behaves as if no buffer were present Acid-Base Titrations ν A titration is a method used to determine the concentration of an unknown species ν Add a measured amount of a known reactant ν Determine when the reaction has gone to completion [unknown] + [known] → products ν At the equivalence point moles unknown = moles known added CunknownVunknown = CknownVknown 7 Acid-Base Titrations Titrate 50.00 10 mL unknown 9 HCl soln. 8 equivalence point with 0.2137 7 M NaOH 6 5 4 3 2 1 0 5 10 15 20 25 30 35 40 mL NaOH added Acid-Base Titrations 10 At equivalence point, VNaOH = 23.96 mL 9 8 equivalence point mol(NaOH) = 7 (.2137 M)(.02396 L) 6 = 5.120 x 10-3 mol 5 4 -3 mol(HCl) = 5.120 x 10 mol 3 (mol known = mol unknown) 2 1 [HCl] = 0 5 10 15 20 25 30 35 40 (5.120x10-3 mol)/(.05000 L) mL NaOH added = 0.1024 M 8 Indicators ν An indicator is a chemical species that changes color depending on the pH of the solution ν An indicator is a conjugate acid-conjugate base pair in which the acid and base forms of the compound have different colors → - + HIn(aq) + H2O In (aq) + H3O (aq) color 1 color 2 ν Indicators are used to determine the endpoint of a titration Indicators ν The pKa of the indicator determines the pH range over which the color changes [HIn]/[In-] ≥ 10 acid color [HIn]/[In-] ″ 0.1 base color [HIn]/[In-] ≈ 1 intermediate color ν - Remember: pH = pKa + log{[In ]/[HIn]} If [HIn]/[In-] = 1, log{[HIn]/[In-]} = 0 ∴ pH = pKa at point when indicator is changing color 9 Indicators Indicator pKa pH range color change Methyl orange 3.7 3.1 – 4.4 red to yellow Bromophenol blue 4.0 3.0 – 4.6 yellow to blue Methyl red 5.1 4.2 – 6.3 red to yellow Bromothymol blue 7.0 6.0 – 7.6 yellow to blue Phenol red 7.9 6.8 – 8.4 yellow to red Phenolphthalein 9.3 8.2 – 10.0 clear to pink Indicators Figure 17.5: pH curve for change after endpoint titration of 0.100 M HCl with 0.100 M NaOH change around endpoint change before endpoint 10 Indicators ν Titration of weak acid with strong base - → - HA(aq) + OH (aq) A (aq) + H2O ν At equivalence point - ↔ - A (aq) + H2O HA(aq) + OH (aq) the solution is basic because conjugate base of weak acid reacts with water to form OH-(aq) Indicators ν Titrate 25.00 mL 0.100 M formic acid (HCOOH) with 0.100 M NaOH -4 Ka = 1.8 x 10 ν Find pH at equivalence point and select appropriate indicator ν At equivalence point, mol(HCOOH) = mol(OH-) mol(fa) = (0.100 M fa)(0.02500 L) = 2.5 x 10-3 mol fa = formic acid -3 VNaOH added = (2.5 x 10 mol)/(0.100 M) = 25.0 mL Vtotal = 50.0 mL 11 Indicators ν Assume HCOOH + OH- reaction goes to completion: [HCOO-] = (2.5 x 10-3 mol)/(0.0500 L) = 0.0500 M ν Determine Keq for reaction of formate ion: - ↔ - HCOO (aq) + H2O HCOOH(aq) + OH (aq) - = [HCOOH][OH ] = Kw = -11 Keq − 6.67 x 10 [HCOO ] Ka Indicators - = [HCOOH][OH ] = Kw = -11 Keq − 6.67 x 10 [HCOO ] Ka [HCOO-] [HCOOH] [OH-] initial .0500 0 0 -x x x equil .0500 – x x x [HCOOH][OH-] x2 K = = = 6.67 x 10-11 eq [HCOO-] 0.0500 − x 12 Indicators [HCOOH][OH-] x2 K = = = 6.67 x 10-11 eq [HCOO-] 0.0500 − x x = 1.83 x 10-6 M = [OH-] pOH = -log(1.83 x 10-6) =5.74 pH = 14.00 – 5.74 = 8.26 Phenol red (6.8 – 8.4) or phenolphthalein (8.2 – 10.0) would be appropriate indicators Indicators ν Titration of weak base with strong acid + → + B(aq) + H3O (aq) BH (aq) + H2O ν At equivalence point + ↔ + BH (aq) + H2O B(aq) + H3O (aq) the solution is acidic because conjugate acid of + weak base reacts with water to form H3O (aq) 13 Indicators Figure 17.8: titration of 0.100 M NH3 with 0.100 M HCl Acid Rain ν Carbon dioxide in the air is in equilibrium with H2O in atmospheric water droplets (clouds & fog): ↔ CO2(aq) + H2O H2CO3(aq) -7 carbonic acid Ka = 4.2 x 10 ↔ + - H2CO3(aq) + H2O H3O (aq) + HCO3 (aq) ν Natural rain water has pH = 5.6 14 Acid Rain ν Emitted pollutants can form additional acid sources in clouds: NO2: ↔ 2 NO2(aq) + H2O HNO3(aq) + HNO2(aq) nitric acid nitrous acid -4 strong Ka = 4.5 x 10 Acid Rain ν Emitted pollutants can form additional acid sources in clouds: SO2: ↔ SO2(aq) + H2O H2SO3(aq) -2 sulfurous acid Ka = 1.2 x 10 → 2 SO2(g) + O2(g) 2 SO3(g) ↔ SO3(aq) + H2O H2SO4(aq) sulfuric acid strong 15 Acid Rain Solubility Products ν Many salts are only slightly soluble ν The solubility product is a measure of the concentration of ions in a solution saturated with the salt ↔ + - + - MA(s) M (aq) + A (aq) Ksp = [M ][A ] Examples ↔ + - + - -10 AgCl(s) Ag (aq) + Cl (aq) Ksp=[Ag ][Cl ]=1.8x10 ↔ 2+ - 2+ - 2 -5 PbCl2(s) Pb (aq) + 2 Cl (aq) Ksp=[Pb ][Cl ] =1.7x10 ↔ 3+ - 3+ - 3 -36 AuBr3(s) Au (aq) + 3 Br (aq) Ksp=[Au ][Br ] =4.0x10 16 Solubility Products ν Knowing the Ksp, we can calculate the concentration of ions in solution Examples ↔ + - + - -10 AgCl(s) Ag (aq) + Cl (aq) Ksp=[Ag ][Cl ]=1.8x10 °-x x x x2 = 1.8 x 10-10 ⇒ x = 1.3 x 10-5 M = [Ag+] = [Cl-] Solubility Products Examples ↔ 2+ - 2+ - 2 -5 PbCl2(s) Pb (aq) + 2 Cl (aq) Ksp=[Pb ][Cl ] =1.7x10 °-x x 2x x(2x)2 = 1.7 x 10-5 4x3 = 1.7 x 10-5 ⇒ x = 1.6 x 10-2 M [Pb2+] = 1.6 x 10-2 M [Cl-] = 3.2 x 10-2 M 17 Solubility Products Examples ↔ 3+ - 3+ - 3 -36 AuBr3(s) Au (aq) + 3 Br (aq) Ksp=[Au ][Br ] =4.0x10 °-x x 3x x(3x)3 = 4.0 x 10-36 27x4 = 4.0 x 10-36 ⇒ x = 6.2 x 10-10 M [Au3+] = 6.2 x 10-10 M [Br-] = 1.9 x 10-9 M Solubility Products Examples—Common ion effect How much PbI2 will dissolve in a 0.0100 M solution of NaI? ↔ 2+ - -9 PbI2(s) Pb (aq) + 2 I (aq) Ksp = 8.7 x 10 °-x x 2x + .0100 x(2x+.0100)2 = 8.7 x 10-9 x(4x2 + 0.0200x + 1.0x10-4) = 8.7 x 10-9 4x3 + .0200x2 + 1.0x10-4x – 8.7x10-9 = 0 x = 8.6 x 10-5 M vs 1.3 x 10-3 M if no I-(aq) were present initially 18 Factors Affecting Solubility ν Salts that are slightly soluble in water can be much more soluble in acid if one or both of its ions are moderately basic: ↔ 2+ 2- -9 CaCO3(s) Ca (aq) + CO3 (aq) Ksp = 8.7 x 10 ν 2- - But CO3 (aq) is the conjugate base of HCO3 (aq) 2- + ↔ - -4 CO3 (aq) + H3O (aq) HCO3 (aq) + H2OKb = 2.1x10 - + ↔ -8 HCO3 (aq) + H3O (aq) H2CO3(aq) + H2OKb = 2.4x10 ↔ ≈ 5 H2CO3(aq) H2O + CO2(g) Keq 10 ν Works for carbonates, some sulfides, phosphates, etc.
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  • CHAPTER 17: Advanced Acid-Base Equilibria

    CHAPTER 17: Advanced Acid-Base Equilibria

    Chapter 17 Advanced Acid-Base Equilibria SY 4/12/11 CHAPTER 17: Advanced Acid-Base Equilibria Chapter 17 17.1 Acid-Base Reactions 17.2 Buffers 17.3 Acid-Base Titrations Chapter In Context 17.4 Polyprotic Acids We will now expand the introductory coverage of acid-base equilibria in the previous 17.5 Two Important Buffer Systems chapter and explore the chemistry of more complicated aqueous solutions containing acids and bases. First we will address the different types of acid-base reactions and then move on to study buffer solutions, acid-base titrations, and polyprotic acids. So that you can get a feeling for the importance of buffers in your world, we will also briefly discuss Chapter Goals the chemistry of two important buffers in biological systems. In the following chapter Recognize the different we will conclude our coverage of chemical equilibria with Lewis acids and bases and the types of and the extent equilibria of sparingly soluble compounds. of acid-base reactions. Describe the One of the more important types of acid-base solutions in terms of commercial and components of a buffer. biological applications are buffers because they allow us to control the pH of a solution. Apply the principles of Buffers play an important role wherever you look: acid-base equilibria to Biology: You are composed of molecules that depend on hydrogen bonding for their buffer solutions. Apply the concepts of structure and function, and are therefore highly sensitive to pH. Most of the reactions acid-base equilibria to in your body occur in aqueous solutions containing buffering agents.
  • Buffers and the Common-Ion Effect (Ch 16) a Buffer Works Through the Common-Ion Effect

    Buffers and the Common-Ion Effect (Ch 16) a Buffer Works Through the Common-Ion Effect

    Buffers and the Common-ion Effect (Ch 16) A buffer works through the common-ion effect. Acetic acid in water dissociates slightly to produce some acetate ion: - + CH3COOH(aq) + H2O(l) CH3COO (aq) + H3O (aq) acetic acid acetate ion If NaCH3COO is added, it provides a source of - CH3COO ion, and the equilibrium shifts to the left. - CH3COO is common to both solutions. - The addition of CH3COO reduces the % dissociation of the acid. ©2013 McGraw-Hill Ryerson Limited 17-1 Table 17.1 The Effect of Added Acetate Ion on the Dissociation of Acetic Acid - * + [CH3COOH]init [CH3COO ]added % Dissociation [H3O ] pH 0.10 0.00 1.3 1.3x10-3 2.89 0.10 0.050 0.036 3.6x10-5 4.44 0.10 0.10 0.018 1.8x10-5 4.74 0.10 0.15 0.012 1.2x10-5 4.92 [CH COOH] * % Dissociation = 3 dissoc x 100 [CH3COOH]init ©2013 McGraw-Hill Ryerson Limited 17-2 How a Buffer Works The buffer components (HA and A-) are able to consume - + small amounts of added OH or H3O by a shift in equilibrium position. - + CH3COOH(aq) + H2O(l) CH3COO (aq) + H3O (aq) - + Added OH reacts with Added H3O reacts with - CH3COOH, causing a shift to CH3COO , causing a the right. shift to the left. The shift in equilibrium position absorbs the change in + - [H3O ] or [OH ], and the pH changes only slightly. ©2013 McGraw-Hill Ryerson Limited 17-3 Figure 17.3 How a buffer works. Buffer has more HA after Buffer has equal Buffer has more A- after + - - addition of H3O .
  • Buffer Solution Adapted from WIKI

    Buffer Solution Adapted from WIKI

    技 术 资 料 集 锦 NO：20140924 Buffer solution in Biology adapted from WIKI A buffer solution (more precisely, pH buffer or hydrogen ion buffer) is an aqueous solution consisting Acids and bases of a mixture of a weak acid and its conjugate base, or vice versa. Its pH changes very little when a small amount ofstrong acid or base is added to it and thus it is used to prevent changes in the pH of a solution. Buffer solutions are used as a means of keeping pH at a nearly constant value in a wide variety of chemical applications. Many life forms thrive only in a relatively small pH range so they utilize a buffer solution to maintain a constant pH. One example of a buffer solution found in nature is blood. pH Acid–base reaction Contents Acid–base titration Acid-base extraction Dissociation constant 1 Principles of buffering Acid dissociation constant 2 Applications Acid strength o 2.1 Simple buffering agents Acidity function o 2.2 "Universal" buffer mixtures Buffer solutions o 2.3 Common buffer compounds used in biology Proton affinity 3 Buffer capacity Self-ionization of water 4 Calculating buffer pH Amphoterism o 4.1 Monoprotic acids Acid types o 4.2 Polyprotic acids 5 See also Brønsted 6 References Lewis 7 External links Mineral 0510 81819585 | 189 51576701 | QQ: 2642 166682 | [email protected] 1 无锡天演生物技术有限公司 | 江苏省无锡市智慧路 33 号 7 栋 601 室 技 术 资 料 集 锦 NO：20140924 Organic Strong Superacids Principles of buffering Weak Simulated titration of an acidified solution of a weak acid (pKa = Base types 4.7) with alkali.