Acid-Base Equilibria Chapter 17 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Chapter 17: Acid-Base Equilibria I. II. Buffer Composition pH of Buffer Solutions • pH after small amounts of acid/base added III. Designing Buffer Solutions IV. Acid-Base Titrations • Indicator A buffer solution has the ability to resist large changes in pH upon the addition of small amounts of either acid or base. A buffer solution is a solution of: 1. A weak acid and its conjugate base or 2. A weak base and it conjugate acid Note: If either the acid or base is a negative or positive ion; it will be found in the form of a salt. Note: It is important that the acid and base have the conjugate relationship. 16.3 Six Strong Acids HCl H2SO4 HBr HClO4 HI HNO3 Which of the following are buffer systems? (a) KF/HF (b) KBr/HBr, (c) Na2CO3/NaHCO3 (a) F- is a weak base and HF is its weak conjugate acid buffer solution (b) HBr is a strong acid not a buffer solution (c) CO32- is a weak base and HCO3- is it conjugate acid buffer solution 16.3 How a Buffer Withstands Large Changes in pH (CH3COO)H/Na(CH3COO-) How a Buffer Withstands Large Changes in pH (CH3COO)H/Na(CH3COO-) Consider an equal molar mixture of CH3COOH and CH3COONa Add strong acid H+ (aq) + CH3COO- (aq) Add strong base OH- (aq) + CH3COOH (aq) CH3COOH (aq) CH3COO- (aq) + H2O (l) Base of buffer removes excess H+ Acid of buffer removes excess OH- How a Buffer Withstands Large Changes in pH (CH3COO)H/Na(CH3COO-) Introductory Chemistry 2/e by N Tro, Prentice Hall, 2006, pg 495 A buffer solution is the most effective if ... 1) there are large amounts of acid/conjugate base 2) the amounts of acid and conjugate base are about equal To find pH of buffer made of; HF/FHF (aq) <-> H+(aq) + F- (aq) Ka = [H+][F-] [HF] H+ = Ka [HF] [F-] If the ratio ([HA]/[A-]) stays about same; pH will not change dramatically ! A buffer solution is the most effective if ... 1) there are large amounts of acid/conjugate base 2) the amounts of acid and conjugate base are about equal HF F- Add OH- HF F- OH- (aq) + HF (aq) <-> F- (aq) + H2O (l) Free (OH)- reacts with/removes acid (HF) of buffer and produces more F-. Add H+ HF F- H+ (aq) + F-(aq) <-> HF (aq) Free H+ reacts with/removes base (F-) of buffer and produces more HF. A buffer solution is the most effective if ... 1) there are large amounts of acid/conjugate base 2) the amounts of acid and conjugate base are about equal FHF Add 4998 = 0.999 5002 3 = 0.4286 7 OH- HF F- 5000 = 1 5000 5=1 5 Add H+ HF F- 5002 = 1.0008 4998 7 = 2.333 3 Determining the pH of Buffer Solutions • The calculations are very similar to determining the pH of weak acid solutions– the only difference is that initially there are both some reactants and products present. An easier way to determine pH of Buffer solutions Henderson-Hasselbalch equation Consider mixture of salt NaA and weak acid HA. HA (aq) H+ (aq) + A- (aq) [H+][A-] Ka = [HA] [H+] Ka [HA] = [A-] -log [H+] = -log Ka [HA] [A-] [HA] -log [H+] = -log Ka - log [A-] [A-] pH = pKa + log [HA] Henderson-Hasselbalch equation [conjugate base] pH = pKa + log [weak acid] [A-] pH = pKa + log [HA] Insignificant Change Approximation is Assumed In This Equation! What is the pH of a solution containing 0.30 M HCOOH and 0.52 M HCOOK? Mixture of weak acid and conjugate base! HCOOH (aq) Initial (M) Change (M) Equilibrium (M) Common ion effect 0.30 – x 0.30 0.52 + x 0.52 H+ (aq) + HCOO- (aq) 0.30 0.00 0.52 -x +x +x 0.30 - x x 0.52 + x [HCOO-] pH = pKa + log [HCOOH] [0.52] = 4.01 pH = 3.77 + log [0.30] HCOOH pKa = 3.77 16.2 Calculate the pH of the 0.30 M NH3/0.36 M NH4Cl buffer system. What is the pH after the addition of 20.0 mL of 0.050 M NaOH to 80.0 mL of the buffer solution? NH4+ (aq) [NH3] pH = pKa + log [NH4+] start (moles) end (moles) H+ (aq) + NH3 (aq) pKa = 9.25 0.029 0.001 NH4+ (aq) + OH- (aq) 0.028 0.0 [0.30] pH = 9.25 + log = 9.17 [0.36] 0.024 H2O (l) + NH3 (aq) 0.025 final volume = 80.0 mL + 20.0 mL = 100 mL [NH4 +] 0.028 0.025 = [NH3] = 0.10 0.10 [0.25] pH = 9.25 + log = 9.20 [0.28] 16.3 Designing Buffer Solutions pH = pKa + log {[X-]/[HX]} • Buffers are most effective if there are equal and large amounts of weak acid and conjugate base. [X-] = [HX] • pH = pKa + log (1) • pH = pKa pKa of Weak Acids Weak Acid pKa Hydrofluoric Acid 3.15 Benzoic Acid 4.19 Carbonic Acid 6.38 Hydrocyanic Acid 9.31 Titrations In a titration a solution of accurately known concentration is added gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete. Equivalence point – the point at which the reaction is complete Indicator – substance that changes color at (or near) the equivalence point Slowly add base to unknown acid UNTIL The indicator changes color (pink) 4.7 Acid-Base Titrations Introductory Chemistry 2/e by N Tro, Prentice Hall, 2006, pg 480 At endpoint/ equivalence point + moles H = moles (OH) MV(acid) = MV(base) • Two ways to get endpoint/equivalence point 1. Color change 2. Mid-point of steep rise of acid/base titration curve Strong Acid-Strong Base Titrations NaOH (aq) + HCl (aq) H2O (l) + NaCl (aq) neutral salt 0.10 M NaOH added to 25 mL of 0.10 M HCl 16.4 Weak Acid-Strong Base Titrations CH3COOH (aq) + NaOH (aq) CH3COONa (aq) + H2O (l) Basic Salt At equivalence point (pH > 7): 16.4 Strong Acid-Weak Base Titrations HCl (aq) + NH3 (aq) NH4Cl (aq) Acidic Salt At equivalence point (pH < 7): 16.4 The titration curve of a strong acid with a strong base. 16.5 Which indicator(s) would you use for a titration of HNO2 with KOH ? Weak acid titrated with strong base. At equivalence point, will have conjugate base of weak acid. At equivalence point, pH > 7 Use cresol red or phenolphthalein 16.5