1. Ch. 15 & 16 Review Everything except Polyprotics & Lewis Acids/Bases!!

2. Practice 1 Write the equilibrium-constant expressions for the following processes: a) H 2(g) + I 2(g)  2HI (g) b) Cd 4+ (aq) + 4Br - (aq)  CdBr 4(aq) c) P 4 (s) + 5 O 2 (g) ↔ P 4 O 10 (s)

3. Practice 2 For the UNBALANCED reaction: NH 3 (aq) ↔ N 2 (g) + H 2 (g) 1. Write an equilibrium expression 2. Calculate the value of K at 127˚C for:  [NH 3 ] = 3.1x10 -2 M  [N 2 ] = 8.5x10 -1 M  [H 2 ] = 3.1x10 -3 M 3. Calculate the value of K with the above concentrations for the following reaction: 1/2 N 2 + 3/2 H 2 ↔ NH 3

4. Practice 3 A mixture of hydrogen and nitrogen in a reaction vessel is allowed to attain equilibrium at 472°C. The equilibrium mixture of gases was analyzed and found to contain 7.38 atm H 2, 2.46 atm N 2, and 0.166 atm NH 3. From these data, calculate the equilibrium constant K p for the reaction

5. Practice 4 Sulfur trioxide decomposes at high temperature in a sealed container: Initially, the vessel is charged at 1000 K with SO 3(g) at a partial pressure of 0.500 atm. At equilibrium the SO 3 partial pressure is 0.200 atm. Calculate the value of K p at 1000 K.

6. Practice 5 At 448°C the equilibrium constant K c for the reaction is 50.5. Predict in which direction the reaction will proceed to reach equilibrium at 448°C if we start with 2.0  10 –2 mol of HI, 1.0  10 –2 mol of H 2, and 3.0  10 –2 mol of I 2 in a 2.00-L container.

7. Practice 6 Consider the equilibrium In which direction will the equilibrium shift when (a) N 2 O 4 is added (b) NO 2 is removed (c) the total pressure is increased by addition of N 2(g) (d) the volume is increased (e) the temperature is decreased

8. 7.) Determine if the following is acidic, basic, or neutral. Cu(NO 3 ) 2

9. Answer Ions = Cu +2, NO 3 - Cu +2 + H 2 O ↔ Cu(OH) 2 + H + weak base **acidic (due to H+) NO 3 - + H 2 O ↔ HNO 3 + OH - strong acid **neutral (will dissociate, so H+ and OH- will form water) Salt = Acidic

10. 8.) Determine if the following is acidic, basic, or neutral. KClO 4

11. Answer Ions = K +, ClO 4 - K + + H 2 O ↔ KOH + H + strong base **neutral (will dissociate, so H+ and OH- will form water) ClO 4 - + H 2 O ↔ HClO 4 + OH - strong acid **neutral (will dissociate, so H+ and OH- will form water) Salt = neutral

12. 9.) Determine if the following is acidic, basic, or neutral. NaH 2 PO 4

13. Answer Ions = Na +, H 2 PO 4 - Na + + H 2 O ↔ NaOH + H + strong base **neutral (will dissociate, so H+ and OH- will form water) H 2 PO 4 - + H 2 O ↔ H 3 PO 4 + OH - weak acid **basic (due to OH-) Salt = basic

14. 10.) Determine if the following is acidic, basic, or neutral. LiF

15. Answer Ions = Li +, F - Li + + H 2 O ↔ LiOH + H + strong base **neutral (will dissociate, so H+ and OH- will form water) F - + H 2 O ↔ HF + OH - weak acid **basic (due to OH-) Salt = basic

16. 11) Determine if the following is acidic, basic, or neutral. (NH 4 ) 2 CO 3 Extra info:  K b (of NH 3 ) = 1.8x10 -5  k a (of HCO 3 - ) = 5.6x10 -11

17. Answer Ions = NH 4 +, CO 3 -2 NH 4 + + H 2 O ↔ NH 3 + H 3 O + weak base **acidic (due to H 3 O + ) CO 3 - + H 2 O ↔ HCO 3 - + OH - weak acid **basic (due to OH-) **must compare K b vs. K a to decide pH of salt K b (of NH 3 ) = 1.8x10 -5 K a of NH 4 + = 5.6x10 -10 k a (of HCO 3 - ) = 5.6x10 -11 k b of CO 3 -2 = 1.8x10 -4 Salt = basic since K b is greater than k a

18. 12.) Complete the following table: pH[H+]pOH[OH–]Acidic, basic, or neutral? 5.4 x 10 –4 7.8 x 10 -10 10.75 5.00

19. Answers: ROW1 pH = 3.27 pOH = 10.73 [OH–] = 1.9 x 10 –11 acidic (since pH < 7) ROW 2 pH = 4.89 [H+] = 1.3 x 10 –5 pOH = 9.11 acidic (since pH < 7) ROW 3 [H+] = 1.8 x 10 -11 pOH = 3.25 [OH–] = 5.6 x 10 –4 basic (since pH > 7) ROW 4 pH = 9.00 [H+] = 1.0 x 10 –9 [OH–] = 1.0 x 10 –5 basic (since pH > 7)

20. Practice 13 Calculate the pH of a 0.0430 M HNO 3 solution.

21. Answer Since HNO 3 is a strong acid, the nitric acid solution will be 100% ionized. Thus [H+] = [NO 3 – ] = 0.0430 M. The pH = - log [0.0430] = 1.37

22. Practice 14 Calculate the pH of a 0.020 M Ba(OH) 2 (aq) solution.

23. Answer Since Ba(OH) 2 is a strong base it is 100% ionized. Note that ionization gives 2 OH – ions for each mole of Ba(OH) 2. Thus [OH – ] = 2 x 0.020 M = 0.040 M pOH = -log[0.040] = 1.40 pH = 14 – 1.40 = 12.60

24. Practice 15 Calculate the pH of a 0.250 M HC 2 H 3 O 2 solution. K a (HC 2 H 3 O 2 ) = 1.8 x 10 -5.

25. Answer Answer: Balanced Equation HC 2 H 3 O 2  H + + C 2 H 3 O 2 – Initial Concentration (M)0.25000 Change (M)- xxx Equilibrium Concentration (M)0.250 - xxx (a) Thus x 2 = 4.5 x 10 -6 ; x = 2.12 x 10 -3 = [H + ]. pH = 2.67.

26. Practice 16 Calculate the pH of a 0.600 M solution of methylamine CH 3 NH 2. K b = 4.4 x 10 –4.

27. Answer Answer: Since CH 3 NH 2 is a weak base, the balanced equation for the reaction is CH 3 NH 2 + H 2 O  CH 3 NH 3 + + OH –. Balanced EquationCH 3 NH 2    CH 3 NH 3 + + OH – Initial Concentration (M)0.600 ___ 00 Change (M)- x ___ xx Equilibrium Concentration (M)0.600 - x ___ xx Thus x = 1.62 x 10 -2 = [OH – ], and pOH = 1.79. The pH = 12.21.

28. Practice 17 The pH of a 0.10 M solution of a weak base is 9.67. What is the K b of the base?

29. Answer Answer: The balanced equation for a weak base B is given in Eq(10). The equilibrium table required is given below. Balanced EquationB    BH + + OH – Initial Concentration (M)0.10 ___ 00 Change (M)- x ___ xx Equilibrium Concentration (M)0.10 - x ___ xx At equilibrium, [OH – ] = [BH + ] = x. Use the pH to calculate the [OH – ] at equilibrium (which is the value of x). Here pOH = 14.00 – pH = 14.00 – 9.67 = 4.33. Thus :

30. Practice 18 Use the following acidity constants to help answer the questions below:  K a (HC 2 H 3 O 2 ) = 1.8 x 10 – 5  K a (HCN) = 4.9 x 10 –10  Ka(HCOOH) = 1.7 x 10 -4 1. Which of the three acids is the weakest? 2. Which of the following bases is the strongest: C 2 H 3 O 2 -, CN -, or HCOO - ? 3. What is the pK a of HCN? 4. What is the K b for CN - ?

31. Answer (1) smallest k value = HCN (2) strong base = weakest acid = lowest k value = HCN (acid) = CN - (base) (3) pK a = -log K a = -log (4.9x10 -10 ) = 9.31 (4) Ka x Kb = 1x10 -14 k b = 1x10 -14 = 2.0 x 10 -5 4.9x10 -10

32. 19.) Predict whether an aqueous solution would be acidic, basic or neutral? 1.sodium nitrate 2.ammonium iodide 3.sodium bicarbonate 4.ammonium cyanide 5.sodium hypochlorite 6.potassium acetate

33. Predict whether an aqueous solution would be acidic, basic or neutral? ANSWERS 1.sodium nitrate = NaNO 3 = neutral 2.ammonium iodide = NH 4 I = acidic 3.sodium bicarbonate = NaHCO 3 = basic 4.ammonium cyanide = NH 4 CN = basic (K b ↑) k b (for NH 3 ) vs. K a (for HCN) 1.8x10 -5 vs. 4.9x10 -10 5.sodium hypochlorite = NaClO = basic 6.potassium acetate = KC 2 H 3 O 2 = basic

34. Practice 20 (from ch. 15 study questions) Suppose that 0.50 moles of hydrogen gas, 0.50 moles of iodine gas, and 0.75 moles of hydrogen iodide gas are introduced into a 2.0 Liter vessel and the system is allowed to reach equilibrium. H 2 (g) + I 2 (g) ↔ 2 HI(g) Calculate the concentrations of all three substances at equilibrium. At the temperature of the experiment, K c equals 2.0 x 10 -2.

35. Practice 21 (from ch. 15 study questions) Nitrosyl chloride NOCl decomposes to nitric oxide and chlorine when heated: 2 NOCl(g) ↔ 2 NO(g) + Cl 2 (g) At 600K, the equilibrium constant K p is 0.060. In a vessel at 600K, there is a mixture of all three gases. The partial pressure of NOCl is 675 torr, the partial pressure of NO is 43 torr and the partial pressure of chlorine is 23 torr. a.What is the value of the reaction quotient? b.Is the mixture at equilibrium? c.In which direction will the system move to reach equilibrium? d.When the system reaches equilibrium, what will be the partial pressures of the components in the system? (just set up the problem, do not solve all the way!!)