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Chapter 16 Lecture Notes 7/27/12 Buffer Solutions Buffer Solutions pH of solution adding 0.10 M HCl to 100 mL water n A buffer helps a solution maintain its pH when HCl added pH acid or base is added 7 0 mL 7.00 n A buffer must contain two components to work 6 2 mL 2.71 n a weak acid that reacts with added base 5 n a weak base that reacts with added acid 5 mL 2.32 pH 4 n Buffers usually contain approximately equal 10 mL 2.04 3 amounts of a weak acid and its conjugate base 20 mL 1.78 2 50 mL 1.48 1 0 10 20 30 40 50 mL of 0.10 M HCl added Buffer Solutions Buffer Solutions Solution that is 0.100 M CH3COOH (acetic acid) Find pH of buffer solution: - + and 0.100 M NaCH3COO (sodium acetate) CH3COOH(aq) + H2O ↔ CH3COO (aq) + H3O (aq) Find pH of buffer solution: - + - + [CH3COO ][H3O ] (.100 + x)x -5 CH3COOH(aq) + H2O ↔ CH3COO (aq) + H3O (aq) Ka = = = 1.8 x 10 [CH3COOH] (.100 - x) assume x is negligible - + [CH3COOH] [CH3COO ] [H3O ] compared to .100 M initial 0.100 0.100 ≈0 x = 1.80 x 10-5 M Δ -x x x pH = 4.74 equil 0.100 – x 0.100 + x x Buffer Solutions Buffer Solutions Add 5 mL .10 M HCl Now let solution come to equilibrium - + Find pH of resulting solution [CH3COOH] [CH3COO ] [H3O ] Assume all acid added reacts with acetate ion to form initial .105 .095 ≈0 acetic acid (remember that acids react with bases) Δ -x x x - + CH3COOH(aq) + H2O ↔ CH3COO (aq) + H3O (aq) equil .105 – x .095 + x x [H O+] added = (5 mL)(.10 M)/(105 mL) = 4.76x10-3 M 3 - + [CH COOH] = 0.100 + 0.005 = 0.105 M [CH3COO ][H3O ] (.095 + x)x -5 3 Ka = = = 1.8 x 10 - [CH COOH] (.105 - x) [CH3COO ] = 0.100 - 0.005 = 0.095 M 3 x = 1.99 x 10-5 M pH = 4.71 1 7/27/12 Buffer Solutions Buffer Solutions pH of buffered solution adding 0.10 M HCl to 100 n Henderson-Hasselbach Equation mL soln n Allows calculation of pH of a buffer if 7 HCl added pH concentrations of conjugate acid and conjugate 6 0 mL 4.74 base are known 5 Buffered solution + - 5 mL 4.71 HA(aq) + H2O ↔ H3O (aq) + A (aq) pH 4 10 mL 4.66 + - [H3O ][A ] 3 K = 15 mL 4.58 a [HA] 2 25 mL 4.57 + Ka[HA] [H O ] 1 3 = - 50 mL 4.45 0 10 20 30 40 50 [A ] mL of 0.10 M HCl added Buffer Solutions Buffer Solutions n Take -log of both sides n Using the Henderson-Hasselbach Eqn, we can: n Determine pH of a solution + ⎧Ka[HA]⎫ ⎛[HA]⎞ − log[H3O ] = - log⎨ ⎬ = - log(Ka ) - log⎜ ⎟ - ⎜ - ⎟ n Determine ratio of conjugate base to ⎩ [A ] ⎭ ⎝ [A ] ⎠ conjugate acid to achieve specific pH - ⎛[HA]⎞ ⎛ [A ] ⎞ - -log(Ka) = pKa − log⎜ ⎟ = log⎜ ⎟ ⎛ [A ] ⎞ ⎜[A− ]⎟ ⎜[HA] ⎟ pH = pK + log⎜ ⎟ ⎝ ⎠ ⎝ ⎠ a ⎜[HA]⎟ ⎛ [A- ] ⎞ ⎝ ⎠ pH = pK + log⎜ ⎟ Henderson-Hasselbach Eqn a ⎜[HA]⎟ ⎝ ⎠ Buffer Solutions Buffer Solutions - n Let’s go back to problem of adding HCl to buffer VHCl [HCl] [CH3COOH] [CH3COO ] pH solution: 5mL (.1)(5mL)/105mL .100+.005 .100-.005 = .00476 M =.105 M =.095 M 4.70 n We can use H-H eqn. to make the calculations much easier 10mL (.1)(10)/110 .100+.009 .100-.009 = .00909 M =.109 M =.091 M 4.66 [CH COOH] = 0.100 + [HCl] 3 added 25mL (.1)(25)/125 .100+.020 .100-.020 - [CH3COO ] = 0.100 – [HCl]added = .0200 M =.120 M =.080 M 4.56 50mL (.1)(50)/150 .100+.033 .100-.033 ⎛ [A- ] ⎞ pH = pK + log⎜ ⎟ = .0333 M =.133 M =.067 M 4.44 a ⎜[HA]⎟ ⎝ ⎠ 2 7/27/12 Buffer Solutions Acid-Base Titrations n Buffer Capacity—the amount of acid or base that n A titration is a method used to determine the can be added to a buffer without the pH concentration of an unknown species significantly changing n Add a measured amount of a known reactant n Suppose we acid to a buffer solution: n Determine when the reaction has gone to n The acid will react with the conjugate base until it is completion depleted [unknown] + [known] → products n Past this point, the solution behaves as if no buffer were present Acid-Base Titrations Acid-Base Titrations n For the generic titration reaction n For the generic titration reaction αHA + βB → products n At the equivalence point At the equivalence point moles unknown = moles known added 1 1 C V = C V moles HA = moles base unknown unknown known known α β 1 1 [HA] VHA = [B] VB α β Acid-Base Titrations Acid-Base Titrations 10 10 Titrate 50.00 mL At equivalence point, VNaOH 9 unknown HCl 9 = 23.96 mL 8 equivalence point soln. with 8 equivalence point mol(NaOH) = 7 0.2137 M 7 6 (.2137 M)(.02396 L) NaOH pH 5 6 -3 = 5.120 x 10 mol 4 The titration pH 5 -3 requires 23.96 mol(HCl) = 5.120 x 10 mol 3 4 2 mL of NaOH (mol known = mol unknown) 3 1 to reach the [HCl] = 0 5 10 15 20 25 30 35 40 2 mL NaOH added equivalence (5.120x10-3 mol)/(.05000 L) point 1 0 5 10 15 20 25 30 35 40 = 0.1024 M mL NaOH added 3 7/27/12 Indicators Indicators n An indicator is a chemical species that changes n The pKa of the indicator determines the pH color depending on the pH of the solution range over which the color changes - n An indicator is a conjugate acid-conjugate base [HIn]/[In ] ≥ 10 acid color pair in which the acid and base forms of the [HIn]/[In-] ≤ 0.1 base color compound have different colors [HIn]/[In-] ≈ 1 intermediate color - + HIn(aq) + H2O → In (aq) + H3O (aq) - n Remember: pH = pKa + log{[In ]/[HIn]} color 1 color 2 - - If [HIn]/[In ] = 1, log{[HIn]/[In ]} = 0 n Indicators are used to determine the endpoint pH = pK at point when indicator is of a titration ∴ a changing color Indicators Indicators Indicator pKa pH range color change Methyl orange 3.7 3.1 – 4.4 red to yellow Figure 17.5: pH curve for change after endpoint Bromophenol blue 4.0 3.0 – 4.6 yellow to blue titration of 0.100 M HCl with 0.100 M NaOH change Methyl red 5.1 4.2 – 6.3 red to yellow around endpoint Bromothymol blue 7.0 6.0 – 7.6 yellow to blue Phenol red 7.9 6.8 – 8.4 yellow to red change before endpoint Phenolphthalein 9.3 8.2 – 10.0 clear to pink Indicators Indicators n Titration of weak acid with strong base n Titrate 25.00 mL 0.100 M formic acid (HCOOH) with - - 0.100 M NaOH HA(aq) + OH (aq) A (aq) + H O → 2 -4 Ka = 1.8 x 10 n At equivalence point n Find pH at equivalence point and select appropriate - - A (aq) + H2O ↔ HA(aq) + OH (aq) indicator - the solution is basic because conjugate base of n At equivalence point, mol(HCOOH) = mol(OH ) weak acid reacts with water to form OH-(aq) 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 4 7/27/12 Indicators Indicators - - n Assume HCOOH + OH reaction goes to completion: [HCOOH][OH ] K K = = w = 6.67 x 10-11 - -3 eq − [HCOO ] = (2.5 x 10 mol)/(0.0500 L) = 0.0500 M [HCOO ] Ka n Determine K for reaction of formate ion: eq [HCOO-] [HCOOH] [OH-] HCOO-(aq) + H O ↔ HCOOH(aq) + OH-(aq) 2 initial .0500 0 0 [HCOOH][OH-] K K = = w = 6.67 x 10-11 Δ -x x x eq − [HCOO ] Ka equil .0500 – x x x - 2 [HCOOH][OH ] x -11 Keq = = = 6.67 x 10 [HCOO-] 0.0500 − x Indicators Indicators - 2 [HCOOH][OH ] x n Titration of weak base with strong acid K = = = 6.67 x 10-11 eq - + + [HCOO ] 0.0500 − x B(aq) + H3O (aq) → BH (aq) + H2O x = 1.83 x 10-6 M = [OH-] n At equivalence point pOH = -log(1.83 x 10-6) =5.74 + + BH (aq) + H2O ↔ B(aq) + H3O (aq) pH = 14.00 – 5.74 = 8.26 the solution is acidic because conjugate acid of Phenol red (6.8 – 8.4) or phenolphthalein (8.2 – 10.0) + weak base reacts with water to form H3O (aq) would be appropriate indicators Indicators Acid Rain n 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 + - Figure 17.8: titration of 0.100 M H2CO3(aq) + H2O ↔ H3O (aq) + HCO3 (aq) NH3 with 0.100 M HCl n Natural rain water has pH = 5.6 5 7/27/12 Acid Rain Acid Rain n Emitted pollutants can form additional acid n Emitted pollutants can form additional acid sources in clouds: sources in clouds: NO2: SO2: 2 NO2(aq) + H2O ↔ HNO3(aq) + HNO2(aq) SO2(aq) + H2O ↔ H2SO3(aq) -2 nitric acid nitrous acid sulfurous acid Ka = 1.2 x 10 strong K = 4.5 x 10-4 a 2 SO2(g) + O2(g) → 2 SO3(g) SO3(aq) + H2O ↔ H2SO4(aq) sulfuric acid strong Acid Rain Acid Rain Sulfur content of coal by percentage in the northeastern U.S.
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