1. Acid-Base Equilibria Chapter 16

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. 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

3. 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 CH 3 COOH (aq) H + (aq) + CH 3 COO - (aq) Consider an equal molar mixture of CH 3 COOH and CH 3 COONa Adding more acid creates a shift left IF enough acetate ions are present

4. Which of the following are buffer systems? (a) KF/HF (b) KCl/HCl, (c) Na 2 CO 3 /NaHCO 3 (a) KF is a weak acid and F - is its conjugate base buffer solution (b) HCl 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

5. 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 16.2 Mixture of weak acid and conjugate base! K a for HCOOH = 1.8 x 10 -4 [H + ] [HCOO - ] K a = [HCOOH] x = 1.038 X 10 -4 pH = 3.98

6. OR…… Use the Henderson-Hasselbach equation 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-Hasselbach equation 16.2 pH = pK a + log [conjugate base] [acid]

7. 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!

8. HCl H + + Cl - HCl + CH 3 COO - CH 3 COOH + Cl - 16.3

9. pH= 9.18 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 3 (aq) + H 2 O (l) NH 4 + (aq) + OH - (aq) Initial End 0.30 0.360 0.30 - x0.36 + xx 16.3 [NH 4 + ] [OH - ] [NH 3 ] K b = Change- x + x = 1.8 X 10 -5 (.36 + x)(x) (.30 – x) 1.8 X 10 -5 = 1.8 X 10 -5  0.36x 0.30 x = 1.5 X 10 -5 pOH = 4.82

10. pH = 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) + OH - (aq) H 2 O (l) + NH 3 (aq) start (M) end (M) 0.29 0.01 0.24 0.280.00.25 final volume = 80.0 mL + 20.0 mL = 100 mL 16.3 NH 4 + 0.36 M x 0.080 L = 0.029 mol /.1 L = 0.29 M OH - 0.050 x 0.020 L = 0.001 mol /.1 L = 0.01M NH 3 0.30 M x 0.080 = 0.024 mol /.1 L = 0.24M K a = = 5.6 X 10 -10 [H + ] [NH 3 ] [NH 4 + ] = 5.6 X 10 -10 [H + ] 0.25 0.28 [H + ] = 6.27 X 10 -10

11. = 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 (M) end (M) 0.29 0.01 0.24 0.280.00.25 pH = 9.25 + log [0.25] [0.28] final volume = 80.0 mL + 20.0 mL = 100 mL 16.3

12. Chemistry In Action: Maintaining the pH of Blood 16.3

13. 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 the endpoint (hopefully close to the equivalence point) Slowly add base to unknown acid UNTIL The indicator changes color (pink) 4.7

14. Strong Acid-Strong Base Titrations NaOH (aq) + HCl (aq) H 2 O (l) + NaCl (aq) OH - (aq) + H + (aq) H 2 O (l) 16.4 100% ionization! No equilibrium

15. Weak Acid-Strong Base Titrations CH 3 COOH (aq) + NaOH (aq) CH 3 COONa (aq) + H 2 O (l) CH 3 COOH (aq) + OH - (aq) CH 3 COO - (aq) + H 2 O (l) CH 3 COO - (aq) + H 2 O (l) OH - (aq) + CH 3 COOH (aq) At equivalence point (pH > 7): 16.4

16. Strong Acid-Weak Base Titrations HCl (aq) + NH 3 (aq) NH 4 Cl (aq) NH 4 + (aq) + H 2 O (l) NH 3 (aq) + H + (aq) At equivalence point (pH < 7): 16.4 H + (aq) + NH 3 (aq) NH 4 Cl (aq)

17. Acid-Base Indicators 16.5

18. pH 16.5

19. The titration curve of a strong acid with a strong base. 16.5

20. Which indicator(s) would you use for a titration of HNO 2 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

21. Finding the Equivalence Point (calculation method) Strong Acid vs. Strong Base –100 % ionized! pH = 7 No equilibrium! Weak Acid vs. Strong Base –Acid is neutralized; Need K b for conjugate base equilibrium Strong Acid vs. Weak Base –Base is neutralized; Need K a for conjugate acid equilibrium Weak Acid vs. Weak Base –Depends on the strength of both; could be conjugate acid, conjugate base, or pH 7

22. Exactly 100 mL of 0.10 M HNO 2 are titrated with 100 mL of a 0.10 M NaOH solution. What is the pH at the equivalence point ? HNO 2 (aq) + OH - (aq) NO 2 - (aq) + H 2 O (l) start (moles) end (moles) 0.01 0.0 0.01 NO 2 - (aq) + H 2 O (l) OH - (aq) + HNO 2 (aq) Initial (M) Change (M) Equilibrium (M) 0.050.00 -x-x+x+x 0.05 - x 0.00 +x+x xx [NO 2 - ] = 0.01 0.200 = 0.05 M Final volume = 200 mL K b = [OH - ][HNO 2 ] [NO 2 - ] = x2x2 0.05-x = 2.2 x 10 -11 0.05 – x  0.05x  1.05 x 10 -6 = [OH - ] pOH = 5.98 pH = 14 – pOH = 8.02

23. Complex Ion Equilibria and Solubility A complex ion is an ion containing a central metal cation bonded to one or more molecules or ions. Co 2+ (aq) + 4Cl - (aq) CoCl 4 (aq) 2- Co(H 2 O) 6 2+ CoCl 4 2- 16.10

25. Complex Ion Formation These are usually formed from a transition metal surrounded by ligands (polar molecules or negative ions). As a "rule of thumb" you place twice the number of ligands around an ion as the charge on the ion... example: the dark blue Cu(NH 3 ) 4 2+ (ammonia is used as a test for Cu 2+ ions), and Ag(NH 3 ) 2 +. Memorize the common ligands.

26. Common Ligands LigandsNames used in the ion H2OH2Oaqua NH 3 ammine OH-hydroxy Cl-chloro Br-bromo CN-cyano SCN-thiocyanato (bonded through sulphur) isothiocyanato (bonded through nitrogen)

27. Names Names: ligand first, then cation Examples: –tetraamminecopper(II) ion: Cu(NH 3 ) 4 2+ –diamminesilver(I) ion: Ag(NH 3 ) 2 +. –tetrahydroxyzinc(II) ion: Zn(OH) 4 2- The charge is the sum of the parts (2+) + 4(-1)= -2.

28. When Complexes Form Aluminum also forms complex ions as do some post transitions metals. Ex: Al(H 2 O) 6 3+ Transitional metals, such as Iron, Zinc and Chromium, can form complex ions. The odd complex ion, FeSCN 2+, shows up once in a while Acid-base reactions may change NH 3 into NH 4 + (or vice versa) which will alter its ability to act as a ligand. Visually, a precipitate may go back into solution as a complex ion is formed. For example, Cu 2+ + a little NH 4 OH will form the light blue precipitate, Cu(OH) 2. With excess ammonia, the complex, Cu(NH 3 ) 4 2+, forms. Keywords such as "excess" and "concentrated" of any solution may indicate complex ions. AgNO 3 + HCl forms the white precipitate, AgCl. With excess, concentrated HCl, the complex ion, AgCl 2 -, forms and the solution clears.

29. Coordination Number Total number of bonds from the ligands to the metal atom. Coordination numbers generally range between 2 and 12, with 4 (tetracoordinate) and 6 (hexacoordinate) being the most common.

30. Some Coordination Complexes molecular formula Lewis base/ligand Lewis aciddonor atom coordination number Ag(NH 3 ) 2 + NH 3 Ag + N2 [Zn(CN) 4 ] 2- CN-Zn 2+ C4 [Ni(CN) 4 ] 2- CN-Ni 2+ C4 [PtCl 6 ] 2- Cl-Pt 4+ Cl6 [Ni(NH 3 ) 6 ] 2+ NH 3 Ni 2+ N6