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K >>1 Forward rxn dominates (rxn lies to the right). Mostly products at equilibrium, [products] >> [reactants]

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Presentation on theme: "K >>1 Forward rxn dominates (rxn lies to the right). Mostly products at equilibrium, [products] >> [reactants]"— Presentation transcript:

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5 K >>1 Forward rxn dominates (rxn lies to the right). Mostly products at equilibrium, [products] >> [reactants]

6 K <<1 Reverse rxn dominates (rxn lies to the left). Mostly reactants at equilibrium, [products] << [reactants]

7 K  1 Forward and reverse rxn occur to roughly the same extent, [products]  [reactants]

8 2.0 moles of NH 3 gas are introduced into a previously evacuated 1.0 L container. At a certain temperature the NH 3 partially dissociates by the following equation. At equilibrium 1.0 mol of NH 3 remains. Calculate the equilibrium constant for this reaction.

9 Reaction Quotient (Q) Q = K: The rxn is at equilibrium. No shift. Q < K: The rxn shifts right to produce products to increase Q. Q > K: The rxn shifts left to produce reactants to decrease Q.

10 If a change (stress) is imposed on a system at equilibrium, the position of the equilibrium will shift in a direction that tends to reduce that change (stress). Le Chatelier’s Principle

11 1.SO 2 (g) is removed. 2.O 2 (g) is added. 3.SO 3 (g) is added. 4.The volume of the reaction container is halved. 5.An inert gas like Ar is added. 6.A catalyst is added. 7.Temperature is increased.

12 Le Chatelier’s Principle 1.CO 2 (g) is added. 2.CaCO 3 (s) is added. 3.The volume is increased. 4.The temperature is decreased.

13 1.0 mol of N 2 O 4 (g) is placed in a 10.0 L vessel and then reacts to reach equilibrium. Calculate the equilibrium concentrations of N 2 O 4 and NO 2. K = 4.0 x 10 -7

14 Brønsted-Lowry Model Acids – are proton donors Bases – are proton acceptors

15 HC 2 H 3 O 2 is a stronger acid then HCN which Has the stronger conjugate base?

16 Comments on the Conjugates of Acids and Bases. The weaker the acid the stronger its conjugate base. The weaker the base the stronger its conjugate acid. The conjugate base of a weak acid is a WEAK base. The conjugate base of a strong acid is worthless. The conjugate acid of a weak base is a WEAK acid.

17 AcidKaKa Realative acid strength Conjugate base KbKb Relative base strength HCl~10 6 HF7.2 x HC 2 H 3 O x HOCl3.5 x NH x

18 AcidKaKa Realative acid strength Conjugate base KbKb Relative base strength HCl~10 6 Cl - ~ HF7.2 x F-F- 1.4 x HC 2 H 3 O x C2H3O2-C2H3O x HOCl3.5 x OCl x NH x NH x 10 -5

19 Stuff you should now know. 1.K a value is directly related to acid strength. 2.Weak acids vs. strong acids (K a ’s and % dissociation. 3.Conjugate acid-base pairs. 4.K a K b =K w 5.K b value is directly related to base strength. 6.How to write out K a and K b rxns and expressions. 7.The weaker the acid the stronger the conjugate base (and vice versa). 8.Conjugate bases of strong acids have no basic properties whatsoever! (K b << K w )

20 Calculate the pH of a 0.10 M HBr solution.

21 Calculate the pH of a 0.10 M HOCl solution. K aHOCl = 3.5 x 10 -8

22 Calculate the pH of a 0.10 M NaF. K aHF = 7.2 x 10 -4

23 Calculate the pH of a 0.10 M Ca(OH) 2 solution.

24 Calculate the pH of a solution containing 0.10 M HOCl and 0.02 M NaOCl. K aHOCl = 3.5 x 10 -8

25 Calculate the pOH of 0.05 M Ba(OH) 2.

26 Calculate the pOH of 0.50 M KOCl. K aHOCl = 3.5 x 10 -8

27 Calculate the pOH of 1.00 M HI.

28 Calculate the pOH of 0.25 M NH 4 Cl. K a NH 4 + = 5.6 x

29 Calculate the pOH of a solution containing 0.25 M NH 4 Cl and 0.10 M NH 3. K a NH 4 + = 5.6 x

30 Calculate the pH of 1.6 x M HNO 3.

31 A solution of 8.00 M HCOOH is 0.47% Ionized. What is the K a for the acid? pH?

32 AcidK a HF7.2 x C 6 H 5 NH x HC 2 H 3 O x HCN6.2 x NH 4 +

33 Acidic, Basic, or Neutral? 1.NaCN 2.NH 4 NO 3 3.KI 4.LiC 2 H 3 O 2 5.C 6 H 5 NH 3 Cl 6.KF 7.NaNO 3 8.HClO 4 9.Ca(OH) 2 10.NH 4 CN 11.NH 4 C 2 H 3 O 12.CaO 13.SO 3

34 Acidic, Basic, or Neutral? 1.NaCN Na + - worthless, CN - - weak base, basic 2.NH 4 NO 3 NO worthless, NH weak acid, acidic 3.KI K + - worthless, I - - worthless, neutral 4.LiC 2 H 3 O 2 Li + - worthless, C 2 H 3 O weak base, basic 5.C 6 H 5 NH 3 Cl Cl - - worthless, C 6 H 5 NH weak acid, acidic 6.KF K + - worthless, F - - weak base, basic 7.NaNO 3 Na + - worthless, NO worthless, neutral 8.HClO 4 HClO 4 – strong acid, acidic 9.Ca(OH) 2 Ca(OH) 2 – strong base, basic 10.NH 4 CN K a NH 4 < K b CN - - basic 11.NH 4 C 2 H 3 O K a NH 4 = K b C 2 H 3 O - - neutral 12.CaO metal oxide - basic 13.SO 3 nonmetal oxide - acidic

35 Buffers Buffer – A solution where a weak acid and its conjugate base are both present in solution. Buffers resist changes in pH

36 Good Buffers Good buffers will have the following: –EQUAL concentrations of the weak acid and its conjugate base. –LARGE concentrations of the weak acid and its conjugate base. –pK a = pH of desired pH.

37 Examples of Buffers HCN/CN - NH 4 + /NH 3 H 2 PO 4 - /HPO intracellular fluid buffer H 2 CO 3 /HCO blood buffer

38 Calculate the pH of a solution that is 1.00 M HNO 2 and 1.00 M NaNO 2. K a HNO 2 = 4.0 x 10 -4

39 Calculate the pH when 0.10 mol of HCl is Added to a 1.00 L solution containing 1.00 M HNO 2 and 1.00 M NaNO 2. K a HNO 2 = 4.0 x 10 -4

40 Calculate the pH when 0.10 mol of NaOH are added to a 1.0 L solution containing 1.00 M HNO 2 and 1.00 M NaNO 2. K a HNO 2 = 4.0 x 10 -4

41 Calculate the pH of a solution formed by Mixing mL of M NH 3 and mL of M HCl. K b NH 3 = 1.8 x 10 -5

42 You want to prepare a HOCl buffer of pH You want to make a 500. mL solution and use all of the 0.75 mol of HOCl you have on hand. How many mol of KOCl must you add? K a HOCl = 3.5 x 10 -8

43 Calculate the pH of a solution formed by mixing 500. mL of 1.50 M HCN with 250. mL of 1.00 M NaOH. K a HCN = 6.2 x

44 Total Points in course: 800 Points to be decided next week: ~415

45 Proposed Study Plan Thursday: HE III Material (finish Lon Capa) Friday: HE I Material Saturday: HE II Material Sunday: HE III Material Monday: HE III Material Tuesday: He III Material Wednesday: HE I, II Material Thursday: HE I, II, III Material

46 A 100. mL solution of 0.10 M HF is titrated by 0.10 M NaOH. Calculate the pH when 0.0, 25.0, 50.0, 100.0, and mL of NaOH have been added. K a HF = 7.4 x 10 -4


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