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

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

Buffer Solutions

Buffer Solution

•A solution that resists changes in

pH

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

salt

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

salt

Buffer Examples•Mix acetic acid &

sodium acetate

•Mix ammonia & ammonium chloride

Buffer Solution•A buffer solution

works best when the acid to salt ratio is

1 : 1

Buffer Solution•A buffer solution

works best when the base to salt ratio is

1 : 1

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

pKa

pKa or pKb

•pKa = - log Ka

•pKb = - log Kb

BufferEquilibria

To solve buffer equilibrium

problems, use the same 5 steps

5 Steps of Equilibrium Problems

1) Set up & balance reaction

5 Steps of Equilibrium Problems

2) Assign Equilibrium amounts in terms of x

(ICE)

5 Steps of Equilibrium Problems

3) Write the equilibrium expression (K = ?)

5 Steps of Equilibrium Problems

4) Substitute Equilibrium amounts into the K

5 Steps of Equilibrium Problems

5) Solve for x

Buffer Problems•Calculate the pH of a

solution containing

0.10 M HAc in 0.10 M NaAc: Ka = 1.8 x 10-5

Buffer Problems•Calculate the pH of

0.10 M NH3 in

0.20 M NH4NO3:

•Kb = 1.8 x 10-5

Buffer ProblemsCalculate the pH of a

solution containing

0.10 M HBz in 0.20 M NaBz: Ka = 6.4 x 10-5

Drill:Calculate the pH of a

solution containing

0.30 M HZ in 0.10 M NaZ: Ka = 3.0 x 10-5

Buffer ProblemCalculate the pH of a

solution containing

0.50 M R-NH2 in 0.10 M R-NH3I: Kb = 4.0 x 10-5

Derivations from an

equilibrium constant

HA H+ + A-

[H+][A-]

[HA]Ka =

HA H+ + A-

[H+][A-]

[HA]Ka =

Cross multiply to isolate [H+]

HA H+ + A-

[Ka][HA]

[A-][H+]=

HA H+ + A-

[HA]

[A-][H+] = (Ka)

HA H+ + A-

[HA]

[A-][H+] = (Ka)

Take –log of each side

pH =

[HA]

[A-]pKa - log

Henderson-Hasselbach Eq

[A-]

[HA]pH = pKa + log

Henderson-Hasselbach Eq

[B+]

[B]pOH = pKb+ log

Buffer Problems•Calculate the salt to acid

ratio to make a buffer solution with pH = 5.0

•Ka for HBZ = 2.0 x 10-5

Derivations from an

equilibrium constant

HA H+ + A-

[H+][A-]

[HA]Ka =

[H+][A-]

[HA]Ka =

Divide both sides by [H+]

Ka [A-]

[H+] [HA]=

You Get the Salt to Acid Ratio

Drill:•Calculate the salt to acid

ratio to make a buffer solution with pH = 5.0

•Ka for HBZ = 2.0 x 10-5

Buffer ProblemsCalculate the salt to base

ratio to make a buffer solution with pH = 9.48

•Kb for MOH = 2.0 x 10-5

Equivalence PointPoint at which the # of moles of the two

titrants are equal

Titration Curves

0

2

4

6

8

10

12

14

0.00 10.00 20.00 30.00 40.00 50.00

0

2

4

6

8

10

12

14

0.00 10.00 20.00 30.00 40.00 50.00

[HA]=[A-]

[HA]=[OH-]

Drill:Calculate the pH of a buffer solution containing 0.50 M HX in 0.25 M KX.

Ka = 2.5 x 10-5

0

2

4

6

8

10

12

14

0 20 40 60 80 100

[H2A] = [HA-]

[H2A] = [OH-]

[HA-] = [A-2]

[OH-] = [A-2]

Calculate the HCO3- to

H2CO3 ratio in blood with pH = 7.40

•Ka1 for H2CO3 = 4.4 x 10-7

150 ml of 0.10 M NaOH is added to 100.0 ml of

0.10 M H2CO3. Calculate pH.

•Ka1 for H2CO3 = 4.4 x 10-7

•Ka2 for H2CO3 = 4.8 x 10-11