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Acid-Base Equilibria and Solubility Equilibria Chapter 16 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

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Presentation on theme: "Acid-Base Equilibria and Solubility Equilibria Chapter 16 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display."— Presentation transcript:

1 Acid-Base Equilibria and Solubility Equilibria Chapter 16 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. PowerPoint Lecture Presentation by J. David Robertson University of Missouri

2 The common ion effect is the shift in equilibrium caused by the addition of a compound having an ion in common with the dissolved substance. The presence of a common ion suppresses the ionization of a weak acid or a weak base. (makes the acid or base less soluble) Consider mixture of CH 3 COONa (strong electrolyte) and CH 3 COOH (weak acid). CH 3 COONa (s) Na + (aq) + CH 3 COO - (aq) CH 3 COOH (aq) H + (aq) + CH 3 COO - (aq) common ion 16.2 H + concentration will decrease, and solution will be less acidic

3 Consider mixture of salt NaA and weak acid HA. HA (aq) H + (aq) + A - (aq) NaA (s) Na + (aq) + A - (aq) K a = [H + ][A - ] [HA] [H + ] = K a [HA] [A - ] -log [H + ] = -log K a - log [HA] [A - ] -log [H + ] = -log K a + log [A - ] [HA] pH = pK a + log [A - ] [HA] pK a = -log K a Henderson-Hasselbalch equation 16.2 pH = pK a + log [conjugate base] [acid]

4 What is the pH of a solution containing 0.30 M HCOOH and 0.52 M HCOOK? HCOOH (aq) H + (aq) + HCOO - (aq) Initial (M) Change (M) Equilibrium (M) 0.300.00 -x-x+x+x 0.30 - x 0.52 +x+x x0.52 + x Common ion effect 0.30 – x  0.30 0.52 + x  0.52 pH = pK a + log [HCOO - ] [HCOOH] HCOOH pK a = 3.77 pH = 3.77 + log [0.52] [0.30] = 4.01 16.2 Mixture of weak acid and conjugate base! HCOOH K a = 1.7 x 10 -4

5 A buffer solution is a solution of: 1.A weak acid or a weak base and 2.The salt of the weak acid or weak base Both must be present! A buffer solution has the ability to resist changes in pH upon the addition of small amounts of either acid or base. 16.3 Add strong acid (it reacts with the conjugate base) H + (aq) + CH 3 COO - (aq) CH 3 COOH (aq) Add strong base (it reacts with the acid) OH - (aq) + CH 3 COOH (aq) CH 3 COO - (aq) + H 2 O (l) Consider an equal molar mixture of CH 3 COOH and CH 3 COONa

6 HCl H + + Cl - HCl + CH 3 COO - CH 3 COOH + Cl - 16.3

7 Which of the following are buffer systems? (a) KF/HF (b) KBr/HBr, (c) Na 2 CO 3 /NaHCO 3 (a) KF is a weak acid and F - is its conjugate base buffer solution (b) HBr is a strong acid not a buffer solution (c) CO 3 2- is a weak base and HCO 3 - is it conjugate acid buffer solution 16.3 StrongWeak Anions*OH - Anions in ionic compounds except OH −, Cl −, Br −, I −, NO 3 −, ClO 4 −, HSO 4 −, and H 2 PO 4 − Certain uncharged molecules NoneNH 3 Table: Arrhenius Bases (There are others in each category. You should recognize the examples given.)

8 = 9.20 Calculate the pH of the 0.30 M NH 3 /0.36 M NH 4 Cl 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? NH 4 + (aq) H + (aq) + NH 3 (aq) pH = pK a + log [NH 3 ] [NH 4 + ] pK a = 9.25 pH = 9.25 + log [0.30] [0.36] = 9.17 NH 4 + (aq) + OH - (aq) H 2 O (l) + NH 3 (aq) start (moles) end (moles) 0.0290.001 0.024 0.0280.00.025 pH = 9.25 + log [0.25] [0.28] [NH 4 + ] = 0.028 0.10 final volume = 80.0 mL + 20.0 mL = 100 mL [NH 3 ] = 0.025 0.10 16.3 NH 3 + H 2 O NH 4 + + OH -

9 Chemistry In Action: Maintaining the pH of Blood 16.3


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