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Drill: A 0.100 M solution of HZ ionizes 20.0 %. Calculate: K aHZ.

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Presentation on theme: "Drill: A 0.100 M solution of HZ ionizes 20.0 %. Calculate: K aHZ."— Presentation transcript:

1 Drill: A 0.100 M solution of HZ ionizes 20.0 %. Calculate: K aHZ

2 Buffer Solutions

3 Buffer Solution A solution that resists changes in pH

4 Buffer Solution Made from the combination of a weak acid & its salt

5 Buffer Solution Made from the combination of a weak base & its salt

6 Buffer Examples Mix acetic acid & sodium acetate Mix ammonia & ammonium chloride

7 Buffer Solution A buffer solution works best when the acid to salt ratio is 1 : 1

8 Buffer Solution A buffer solution works best when the base to salt ratio is 1 : 1

9 Buffer Solution The buffering capacity of a solution works best when the pH is near the pK a

10 pK a or pK b pK a = - log K a pK b = - log K b

11 Buffer Equilibria

12 To solve buffer equilibrium problems, use the same 5 steps

13 5 Steps of Equilibrium Problems 1) Set up & balance reaction

14 5 Steps of Equilibrium Problems 2) Assign Equilibrium amounts in terms of x (ICE)

15 5 Steps of Equilibrium Problems 3) Write the equilibrium expression (K = ?)

16 5 Steps of Equilibrium Problems 4) Substitute Equilibrium amounts into the K

17 5 Steps of Equilibrium Problems 5) Solve for x

18 Buffer Problems Calculate the pH of a solution containing 0.10 M HAc in 0.10 M NaAc: K a = 1.8 x 10 -5

19 Buffer Problems Calculate the pH of 0.10 M NH 3 in 0.20 M NH 4 NO 3 : K b = 1.8 x 10 -5

20 Buffer Problems Calculate the pH of a solution containing 0.10 M HBz in 0.20 M NaBz: K a = 6.4 x 10 -5

21 Drill: Calculate the pH of a solution containing 0.30 M HZ in 0.10 M NaZ: K a = 3.0 x 10 -5

22 Buffer Problem Calculate the pH of a solution containing 0.50 M R-NH 2 in 0.10 M R-NH 3 I: K b = 4.0 x 10 -5

23 Derivations from an equilibrium constant

24 HAH + + A - [ H + ][ A - ] [ HA ] K a =

25 HAH + + A - [ H + ][ A - ] [ HA ] K a = Cross multiply to isolate [H + ]

26 HAH + + A - [ K a ][ HA ] [ A - ] [H + ]=

27 HAH + + A - [ HA ] [ A - ] [H + ] = (K a )

28 HAH + + A - [ HA ] [ A - ] [H + ] = (K a ) Take –log of each side

29 pH = [ HA ] [ A - ] pK a - log

30 Henderson- Hasselbach Eq [A - ] [HA] pH = pK a + log

31 Henderson- Hasselbach Eq [B + ] [B] pOH = pK b + log

32 Buffer Problems Calculate the salt to acid ratio to make a buffer solution with pH = 5.0 K a for HBZ = 2.0 x 10 -5

33 Derivations from an equilibrium constant

34 HAH + + A - [ H + ][ A - ] [ HA ] K a =

35 [ H + ][ A - ] [ HA ] K a = Divide both sides by [H + ]

36 K a [ A - ] [ H + ] [ HA ] = You Get the Salt to Acid Ratio

37 Drill: Calculate the salt to acid ratio to make a buffer solution with pH = 5.0 K a for HBZ = 2.0 x 10 -5

38 Buffer Problems Calculate the salt to base ratio to make a buffer solution with pH = 9.48 K b for MOH = 2.0 x 10 -5

39 Equivalence Point Point at which the # of moles of the two titrants are equal

40 Titration Curves

41

42 [HA]=[A - ] [HA]=[OH - ]

43 Drill: Calculate the pH of a buffer solution containing 0.50 M HX in 0.25 M KX. Ka = 2.5 x 10 -5

44

45 Calculate the HCO 3 - to H 2 CO 3 ratio in blood with pH = 7.40 K a1 for H2CO3 = 4.4 x 10 -7

46 150 ml of 0.10 M NaOH is added to 100.0 ml of 0.10 M H 2 CO 3. Calculate pH. K a1 for H2CO3 = 4.4 x 10 -7 K a2 for H2CO3 = 4.8 x 10 -11


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