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.