1. Equilibrium

2. Chemical equilibrium is the state where the concentration of all reactants and products remain constant with time. At equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction. Concentration Time Equilibrium H2H2 NH 3 N2N2

3. Equilibrium constant expression Using the law of mass action, an equilibrium expression can be written: Balanced equation: jA + zB ⇌ lC + mD [C] l [D] m [A] j [B] z K =

4. [C] l [D] m [A] j [B] z K = Equilibrium Expression Equilibrium constant

5. Extent of a Reaction The tendency of a reaction to occur is indicated by the value of K. If K >> 1, at equilibrium the reaction will consist of mostly products; the equilibrium lies to the right. If K << 1, at equilibrium the reaction will consist of mostly reactants; the equilibrium lies to the left.

6. K = [NO] 2 [N 2 ] [O 2 ] Example: Write the equilibrium expression for the followingN 2 (g) + O 2 (g) ⇋ 2NO(g) At the beginning of the experiment, the [N 2 ] = 0.25M and [O 2 ] = 0.43M. Calculate the equilibrium concentration of each substance. The equilibrium constant for the reaction is 4.1 x 10 -4.

7. Initial0.25 M0.43 M0.0 M Change Equilibrium ICE tables N 2 (g) + O 2 (g) ⇌ 2NO(g) - x + 2x 0.25M - x 0.43M - x 2x

8. Substitute into equilibrium expression (2x) 2 (0.25 – x) (0.43 – x) 4.1 x 10 -4 = if we assume x is much, much smaller than 0.25, then the equation becomes easier to deal with. (2x) 2 (0.25) (0.43) 4.1 x 10 -4 = x = 3.3 x 10 -3 M [NO 2 ] = 6.6 x 10 -3 M; [N 2 ] = 0.25 M; [O 2 ] = 0.43 M

9. Example: Calculate the equilibrium concentrations for each substance in the following equation: N 2 O 4 (g) ⇋ 2NO 2 (g) when 0.0150 mol of N 2 O 4 is placed in a 1.0 L container and immediately begins to decompose. The equilibrium constant for the reaction is 0.020. [N 2 O 4 ] = 0.0150 – 0.00866 = 0.00634 M [NO 2 ] = 2x = 2(0.00866) = 0.0173 M

10. Example: Sulfurous acid dissociates in water as follows: H 2 SO 3 (aq) ⇌ H + (aq) + HSO 3 - (aq). If [H 2 SO 3 ] o = 1.50M and [H + ] o = [HSO 3 - ] o = 0.0M, Calculate all of the equilibrium concentrations if K = 1.20 x 10 -2 for this reaction.

11. Example: Calculate the equilibrium concentrations for each substance in a solution of 0.35 M NH 3. The base dissociation constant for NH 3 is 1.8 x 10 -5. NH 3 + HOH ⇋ NH 4 + + OH -

12. Strong Acids HCl HNO 3 H 2 SO 4 HClO 4 Strong Bases LiOHBe(OH) 2 NaOHMg(OH) 2 KOHCa(OH) 2 RbOHSr(OH) 2

13. Answers for Chemquest 50 1.5.6 x 10 -6 10a. [H + ] = 0.0036M 2.5.6 x 10 -10 b. pH = 2.44 3.pOH = 12.5511. pH = 4.69 4.pH = 10.4112. pH = 11.48 5.[H + ] = 0.025M HCl is a strong acid. 6.[OH - ] = 0.75M NaOH is a strong base. 7.pH = 2.31 8.pH = 1.23 pOH = 12.77 9.pH = 12.23

15. Diluting Solutions

16. How many milliliters of aqueous 6.00 M CuSO 4 solution must be diluted with water to prepare 50.0 mL of aqueous 1.00 M CuSO 4 ? 8.33 mL How many milliliters of a solution of 4.00 M KI are needed to prepare 250.0 mL of 0.760 M KI? 47.5 mL

18. A Neutralization Reaction is also called an Acid-Base Reaction Neutralization Reaction Acid + Base  Salt + Water Recall: A ‘salt’ is an ionic compound.

19. Examples: HCl + NaOH  H 2 CO 3 + KOH  2HNO 3 + Ca(OH) 2  NaCl + H 2 O KHCO 3 + H 2 O Ca(NO 3 ) 2 + 2H 2 O

20. How many moles of potassium hydroxide are needed to completely neutralize 1.56 mol of phosphoric acid? How many grams of sodium hydroxide are required to neutralize 0.20 mol of nitric acid?

21. Answers to Skill Practice 50 1.pH = 0.469 2.pH = 2.33 3.Strong acid – K a is large – pH is small Weak acid – K a is small – pH is large 4.K b = 5.88 x 10 -10 5.pH = 2.78 6.K a = 4.04 x 10 -5 7.pH = 11.48 8.K b = 5.58 x 10 -8

22. Answers to Chemquest 51 1.Before: 10. M7a. 0.992 mol HCl, After: 6.2 M 0.984 mol NaOH 2.3.9M b. 0.008 mol HCl 3.1.0M c. 0.0143 M 4.Pour 21.4mL of stock d. pH = 1.84 solution into a 50.0mL8. pH = 13.26 volumetric flask and9. pH = 0.63 fill the flask to the mark10. 8.0x10 -2 mL with distilled water. 5a. 0.3 M b. pH = 0.52 6. 2.5, 1.9

23. Titration: A process used to determine the concentration of a solution in which a solution of a known concentration is added to a measured amount of the solution of unknown concentration until an indicator signals the endpoint. Acid solution with indicator Added base is measured with a buret. Color change shows neutralization.

24. Standardized Solution: A solution where the concentration is KNOWN. In a titration, the endpoint is where the indicator changes color. This will indicate the equivalence point. In a titration, the equivalence point is the point in the reaction when the moles of acid equal the moles of base.

26. Titration of 50 mL of 0.2 M HNO 3 with 0.100 M NaOH.

27. Titration of 100 mL of 0.50 M NaOH with 1.0 M HCl.

28. Titration of 50 mL of 0.100 M acetic acid with 0.100 M NaOH.

29. Comparison of strong and weak acid titration curves.

30. Titration of 100 mL of 0.050 M ammonia with 0.10 M HCl.

31. Titration Calculations: How many milliliters of 0.45M HCl will neutralize 25.0 mL of 1.00M KOH? What is the molarity of H 3 PO 4 if 15.0 mL is completely neutralized by 38.5 mL of 0.150M NaOH?

33. What is different about neutralization reactions that produce a neutral solution at the equivalence point and those that produce acidic or basic solutions? The strengths of the reactants; when a strong acid and base react, the solution is neutral; when a strong base and a weak acid react, the solution is basic; when a strong acid and a weak base react, the solution is acidic.

35. Salts (ionic compounds) can be neutral, acidic, or basic substances. Salt Hydrolysis: When cations or anions of a dissociated salt remove hydrogen ions from water or donate hydrogen ions to water. NH 4 Cl Is Acidic. NaCl Is Neutral. NaC 2 H 3 O 2 Is Basic.

36. In general, salts that produce acidic solutions contain positive ions that release protons to water. Salts that produce basic solutions contain negative ions that attract protons from water.

37. Example: Dissociation of Salt: Positive ion from salt releases protons to water:

38. Example: Dissociation of Salt: Negative Ion from salt attracts proton from water:

39. Buffer A solution of a weak acid and one of its salts, or a solution of a weak base and one of its salts. If an acid is added to a buffer solution, the pH barely changes. If a base is added to a buffer solution, the pH barely changes. Buffer capacity: the amount of acid (or base) that can be added to a buffer solution before a significant change in pH occurs.

40. In buffered aspirin the buffer solution is composed of carbonic acid (H 2 CO 3 ) and hydrogen carbonate ion (HCO 3 - ) Adding a base to a buffer: It reacts with H 2 CO 3 to produce water H 2 CO 3 + OH -  HCO 3 - + H 2 O Adding an acid to a buffer: It reacts with HCO 3 - to produce unionized carbonic acid HCO 3 - + H 3 O +  H 2 CO 3 + H 2 O

41. 1.Write reactions to show what happens when the following occurs a.) Acid is added to a solution of HPO 4 2- b.) Base is added to a solution of H 2 PO 4 - 2. Write an equation that shows what happens when acid is added to the acetic acid (HC 2 H 3 O 2 )/acetate buffer (C 2 H 3 O 2 - ) 3a. Write an equation that shows what happens when acid is added to an ammonium ion (NH 4 + )/ammonia buffer (NH 3 ). 3b. What happens when a base is added?

42. 1a) HPO 4 2- + H 3 O + ⇌ H 2 PO 4 - + H 2 O 1b) H 2 PO 4 + OH - ⇌ HPO 4 2- + H 2 O 2) C 2 H 3 O 2 - + H 3 O + ⇌ HC 2 H 3 O 2 + H 2 O 3a) NH 3 + H 3 O + ⇌ NH 4 + + H 2 O 3b) NH 4 + + OH - ⇌ NH 3 + H 2 O