1. Updates Midterms marked; solutions are posted Assignment 03 is in the box Assignment 04 is up on ACME and is due Mon., Feb. 26 (in class)

2. Acids and Bases Chapter 16

3. How do we measure pH? For less accurate measurements, one can use –Litmus paper Turns blue above ~pH = 8 Turns red below ~pH = 5 –An indicator

4. How do we measure pH? For more accurate measurements, one uses a pH meter, which measures the voltage in the solution.

5. Strong acids You will recall that the seven strong acids are HCl, HBr, HI, HNO 3, H 2 SO 4, HClO 3, and HClO 4. These are, by definition, strong electrolytes and exist totally as ions in aqueous solution. For the monoprotic strong acids, [H 3 O + ] = [acid].

6. Strong bases Strong bases are the soluble hydroxides, which are the alkali metal and heavier alkaline earth metal hydroxides (Ca 2+, Sr 2+, and Ba 2+ ). Again, these substances dissociate completely in aqueous solution.

7. Dissociation constants For a generalized acid dissociation, the equilibrium expression would be This equilibrium constant is called the acid-dissociation constant, K a. [H 3 O + ] [A − ] [HA] K c = HA (aq) + H 2 O (l) A − (aq) + H 3 O + (aq)

8. Dissociation constants The greater the value of K a, the stronger the acid.

9. What is the pH of a 0.5 M HF solution (at 25 0 C)? HF (aq) H + (aq) + F - (aq) K a = [H + ][F - ] [HF] = 7.1 x 10 -4 HF (aq) H + (aq) + F - (aq) Initial (M) Change (M) Equilibrium (M) 0.500.00 -x-x+x+x 0.50 - x 0.00 +x+x xx K a = x2x2 0.50 - x = 7.1 x 10 -4 Ka  Ka  x2x2 0.50 = 7.1 x 10 -4 0.50 – x  0.50 K a << 1 x 2 = 3.55 x 10 -4 x = 0.019 M [H + ] = [F - ] = 0.019 M pH = -log [H + ] = 1.72 [HF] = 0.50 – x = 0.48 M 16.5

10. When can I use the approximation? 0.50 – x  0.50 K a << 1 When x is less than 5% of the value from which it is subtracted. x = 0.019 0.019 M 0.50 M x 100% = 3.8% Less than 5% Approximation ok. What is the pH of a 0.05 M HF solution (at 25 0 C)? Ka  Ka  x2x2 0.05 = 7.1 x 10 -4 x = 0.006 M 0.006 M 0.05 M x 100% = 12% More than 5% Approximation not ok. Must solve for x exactly using quadratic equation. 16.5

11. Solving weak acid ionization problems: 1.Identify the major species that can affect the pH. In most cases, you can ignore the autoionization of water. Ignore [OH - ] because it is determined by [H + ]. 2.Use ICE to express the equilibrium concentrations in terms of single unknown x. 3.Write K a in terms of equilibrium concentrations. Solve for x by the approximation method. If approximation is not valid, solve for x exactly. 4.Calculate concentrations of all species and/or pH of the solution. 16.5

12. What is the pH of a 0.122 M monoprotic acid whose K a is 5.7 x 10 -4 ? HA (aq) H + (aq) + A - (aq) Initial (M) Change (M) Equilibrium (M) 0.1220.00 -x-x+x+x 0.122 - x 0.00 +x+x xx K a = x2x2 0.122 - x = 5.7 x 10 -4 Ka  Ka  x2x2 0.122 = 5.7 x 10 -4 0.122 – x  0.122 K a << 1 x 2 = 6.95 x 10 -5 x = 0.0083 M 0.0083 M 0.122 M x 100% = 6.8% More than 5% Approximation not ok. 16.5

13. K a = x2x2 0.122 - x = 5.7 x 10 -4 x 2 + 0.00057x – 6.95 x 10 -5 = 0 ax 2 + bx + c =0 -b ± b 2 – 4ac  2a2a x = x = 0.0081x = - 0.0081 HA (aq) H + (aq) + A - (aq) Initial (M) Change (M) Equilibrium (M) 0.1220.00 -x-x+x+x 0.122 - x 0.00 +x+x xx [H + ] = x = 0.0081 M pH = -log[H + ] = 2.09 16.5

14. percent ionization = Ionized acid concentration at equilibrium Initial concentration of acid x 100% For a monoprotic acid HA Percent ionization = [H + ] [HA] 0 x 100% [HA] 0 = initial concentration 16.5

15. Calculating K a from the pH The pH of a 0.10 M solution of formic acid, HCOOH, at 25°C is 2.38. Calculate K a for formic acid at this temperature. We know that [H 3 O + ] [COO − ] [HCOOH] K a =

16. Calculating K a from the pH The pH of a 0.10 M solution of formic acid, HCOOH, at 25°C is 2.38. Calculate K a for formic acid at this temperature. To calculate K a, we need the equilibrium concentrations of all three things. We can find [H 3 O + ], which is the same as [HCOO − ], from the pH.

17. Calculating K a from the pH pH = −log [H 3 O + ] 2.38 = −log [H 3 O + ] −2.38 = log [H 3 O + ] 10 −2.38 = [H 3 O + ] 4.2  10 −3 = [H 3 O + ] = [HCOO − ]

18. Calculating K a from pH Now we can set up a table… [HCOOH], M[H 3 O + ], M[HCOO − ], M Initially0.1000 Change −4.2  10 -3 +4.2  10 -3 +4.2  10 −3 At Equilibrium 0.10 − 4.2  10 −3 = 0.0958 = 0.10 4.2  10 −3

19. Calculating K a from pH [4.2  10 −3 ] [0.10] K a = = 1.8  10 −4

20. Polyprotic Acids Have more than one acidic proton. If the difference between the K a for the first dissociation and subsequent K a values is 10 3 or more, the pH generally depends only on the first dissociation.

21. NH 3 (aq) + H 2 O (l) NH 4 + (aq) + OH - (aq) Weak Bases and Base Ionization Constants K b = [NH 4 + ][OH - ] [NH 3 ] K b is the base ionization constant KbKb weak base strength 16.6 Solve weak base problems like weak acids except solve for [OH-] instead of [H + ].

22. 16.6

23. 16.7 Ionization Constants of Conjugate Acid-Base Pairs HA (aq) H + (aq) + A - (aq) A - (aq) + H 2 O (l) OH - (aq) + HA (aq) KaKa KbKb H 2 O (l) H + (aq) + OH - (aq) KwKw K a K b = K w Weak Acid and Its Conjugate Base Ka =Ka = KwKw KbKb Kb =Kb = KwKw KaKa