Acid Acid Equilibrium Equilibrium

and pHand pHSøren Sørensen

Acid/Base Definitions Arrhenius Model

Acids produce hydrogen ions in aqueous solutions

Bases produce hydroxide ions in aqueous solutions

Bronsted-Lowry Model Acids are proton donors Bases are proton acceptors

Lewis Acid Model Acids are electron pair acceptors Bases are electron pair donors

Acid DissociationAcid Dissociation

HA H+ + A-

Acid Proton Conjugate

base

][

]][[

HA

AHKa

Alternately, H+ may be written in its hydrated form, H3O+ (hydronium ion)

Dissociation of Strong AcidsDissociation of Strong AcidsStrong acids are assumed to dissociate completely in solution.

Large Ka or small Ka?Reactant favored or product favored?

Dissociation Constants: Strong Dissociation Constants: Strong AcidsAcids

Dissociation of Weak AcidsDissociation of Weak AcidsWeak acids are assumed to dissociate only slightly (less than 5%) in solution.

Large Ka or small Ka?Reactant favored or product favored?

Dissociation Constants: Weak Dissociation Constants: Weak AcidsAcids

Self-Ionization of WaterSelf-Ionization of Water

H2O + H2O H3O+ + OH-

At 25, [H3O+] = [OH-] = 1 x 10-7

Kw is a constant at 25 C:

Kw = [H3O+][OH-]

Kw = (1 x 10-7)(1 x 10-7) = 1 x 10-14

Calculating pH, Calculating pH, pOHpOHpH = -log10(H3O+)

pOH = -log10(OH-)

Relationship between pH and Relationship between pH and pOHpOH pH + pOH = 14

Finding [HFinding [H33OO++], [OH], [OH--] from pH, pOH] from pH, pOH

[H3O+] = 10-pH

[OH-] = 10-pOH

pH and pOH Calculations

H + O H -

pH pO H

[O H -] = 1 x 10 - 1 4

[H + ]

[H + ] = 1 x 10 - 1 4

[O H -]

p O H = 14 - p H

p H = 14 - p O H

pOH

= -l

og[O

H- ]

pH =

-log

[H+

]

[OH

- ] = 1

0-pO

H

[H+

] = 1

0-pH

The The pH ScalepH Scale

Graphic: Wikimedia Commons user Slower

A Weak Acid Equilibrium A Weak Acid Equilibrium ProblemProblem

What is the pH of a 0.50 M solution of acetic acid, HC2H3O2, Ka = 1.8 x 10-

5 ?Step #1: Write the dissociation equation

HC2H3O2 C2H3O2- + H+

A Weak Acid Equilibrium A Weak Acid Equilibrium ProblemProblem

What is the pH of a 0.50 M solution of acetic acid, HC2H3O2, Ka = 1.8 x 10-

5 ?Step #2: ICE it!

HC2H3O2 C2H3O2- + H+

0.50 0 0

- x +x +x

0.50 - x xx

A Weak Acid Equilibrium A Weak Acid Equilibrium ProblemProblem

What is the pH of a 0.50 M solution of acetic acid, HC2H3O2, Ka = 1.8 x 10-

5 ?Step #3: Set up the law of mass action

HC2H3O2 C2H3O2- + H+

0.50 - x xxEE

)50.0()50.0(

))((108.1

25 x

x

xxx

A Weak Acid Equilibrium A Weak Acid Equilibrium ProblemProblem

What is the pH of a 0.50 M solution of acetic acid, HC2H3O2, Ka = 1.8 x 10-

5 ?Step #4: Solve for x, which is also [H+]

HC2H3O2 C2H3O2- + H+

0.50 - x xxEE

)50.0(108.1

25 x

x [H+] = 3.0 x 10-3 M

A Weak Acid Equilibrium A Weak Acid Equilibrium ProblemProblem

What is the pH of a 0.50 M solution of acetic acid, HC2H3O2, Ka = 1.8 x 10-

5 ?Step #5: Convert [H+] to pH

HC2H3O2 C2H3O2- + H+

0.50 - x xxEE

52.4)100.3log( 5 xpH

• Text Pg. 671 Sample 16.1 and PracticeText Pg. 671 Sample 16.1 and Practice• Sample 16.2 and PracticeSample 16.2 and Practice

• Pg. 673 probably will not appear on AP test.Pg. 673 probably will not appear on AP test.• Equilibrium shifts from the stronger acid to the Equilibrium shifts from the stronger acid to the

stronger base (table will not be available to id stronger base (table will not be available to id the strengths)the strengths)

• Pg. 674 and 675 Sample and practicePg. 674 and 675 Sample and practice• Pg. 677 use Kw to calculate [HPg. 677 use Kw to calculate [H++] or [OH] or [OH--]]• Pg. 677 16.6 and 16.7 Sample and practicePg. 677 16.6 and 16.7 Sample and practice

• Pg. 679 IndicatorsPg. 679 Indicators• Most important phenolphthaleinMost important phenolphthalein• Pink in the presence of a basePink in the presence of a base

• Pg. 680 16.8 and 16.9 Sample and PracticePg. 680 16.8 and 16.9 Sample and Practice

• Pg. 683 Sample and practicePg. 683 Sample and practice

Percent IonizationPercent Ionization

• Percent ionization = Percent ionization = concentration ionizedconcentration ionized X 100 X 100• original concentrationoriginal concentration

• Percent ionization = Percent ionization = [H[H++]]equilibriumequilibrium X 100 X 100

• [HA][HA]initialinitial

• Pg. 684 Sample and practicePg. 684 Sample and practice• Pg. 685 Sample and practicePg. 685 Sample and practice