8.2 – Equilibrium of Weak Acids and Bases

• Acids and bases dissociate in aqueous solutions to form ions that interact with water.

• pH of a solution is determined by the equilibrium of the reaction between water and the ions of the acid or base.

Dissociation of Water

• Pure water contains a few ions from the dissociation of water.

2H20(l) ↔ H3O+(aq) + OH-

(aq)

• At 25°C ~ 2 molecules/1 billion dissociate• Due to the 1:1 ratio of H3O+ to OH- the [H3O+] =

[OH-]• At 25°C, [H3O+] = [OH-] = 1.0x10-7 mol/L

The Ion Product Constant for Water• The equilibrium constant for water

Kc = [H3O+] [OH-]

[H2O]2

• The equilibrium value of [H3O+] [OH-] at 25°C is called the ion constant for water (Kw).

• Kw = [H3O+] [OH-]

= 1.0 x 10-7 mol/L x 1.0 x 10-7 mol/L = 1.0 x 10-14 (units are dropped)

Strong Acids and Bases

• With strong acids and bases, [H3O+] and [OH-] from water is negligible compared to that from the acid or base, so the dissociation of water is ignored.

• Consider 0.1 M HCl,– All of the HCl dissociates, forming [H3O+] = 0.1 mol/L.

2H20(l) ↔ H3O+(aq) + OH-

(aq)

– This forces the equation for dissociation of water to the left (←) (LeChâtelier’s)

– Therefore ,[H3O+] from water is, < 1.0 x 10-7 mol/L– So it can be ignored

[H3O+] and [OH-] at 25°C

• Acidic Solutions[H3O+] > 1.0 x 10-7 mol/L[OH-] < 1.0 x 10-7 mol/L

• Neutral Solutions [OH-] = [H3O+] = 1.0 x 10-7 mol/L

• Basic Solutions[H3O+] < 1.0 x 10-7 mol/L[OH-] > 1.0 x 10-7 mol/L

Determining [H3O+] and [OH-]

• Find [H3O+] and [OH-] in 0.16 M Ba(OH)2

• Ba(OH)2 is a strong base, therefore it dissociates completely.

• Therefore use [Ba(OH)2] to find [OH-] H2O

Ba(OH)2 Ba2+ + 2OH-

0.16 mol/L Ba(OH)2 x 2 mol OH- = 0.32 mol/L OH-

1 mol Ba(OH)2

Determining [H3O+] and [OH-]

• Use Kw = [H3O+] [OH-] = 1.0 x 10-14 to find [H3O+][H3O+][OH-] = 1.0 x 10-14

[H3O+] = 1.0 x 10-14

[OH-] [H3O+] = 1.0 x 10-14 = 3.1 X 10-14 mol/L 0.32 mol/L

Therefore the [H3O+] = 3.1 x 10-14 mol/L, and the[OH-] = 0.32 mol/L

Practice

• Finding [OH-] and [H3O+] for strong acids and bases

• Practice problems on Pg. 537 # 4, 5• Pg. 540 # 10

Calculating pH and pOH

• This should be review from SCH3U• pH = -log[H3O-]

• pOH = -log[OH-]• Kw = [H3O+] [OH-] = 1.0 x 10-14 at 25°C

Therefore, pH + pOH = 14

Problems involving pH and pOH

• A liquid shampoo has a hydroxide ion concentration of 6.8 x 10-5 mol/L at 25°C

a. Is the shampoo acidic, basic or neutral [OH-] = 6.8 x 10-5 mol/L > 1.0 x 10-7 mol/LTherefore, the shampoo is basic

b.Calculate the hydronium ion concentration.[H3O+] = 1.0 x 10-14 = 1.5 x 10-10 mol/L 6.8 x 10-5

Problems involving pH and pOH

• A liquid shampoo has a hydroxide ion concentration of 6.8 x 10-5 mol/L at 25°C

c. What is the pH and the pOH of the shampoo?pH = -log[H3O+] = -log[1.5 x 10-10] = 9.83

pOH = -log[OH-] = -log[6.8 x 10-5] = 4.17

Note: With pH and pOH values. The numbers to the left of the decimal do not count as sig. digits.

Alternative Method for finding [H3O+] and [OH-]

• [H3O+] = 10-pH

• [OH-] = 10-pOH

Ex. If the pH is 5.20, what is the [H3O+] = 10-pH

[H3O+] = 10-pH

[H3O+] = 10-5.2 = 6.3 x 10-6 mol/LPractice Problems• Pg. 546 # 12, 13• Pg. 549 # 17, 18

Acid Dissociation Constant

• Weak acids do not completely dissociate in water.

• For a weak monoprotic acidHA(aq) + H2O(aq) ↔ H3O+

(aq) + A-(aq)

• The equilibrium expression is Kc = [H3O+][A-]

[HA][H2O]

Acid Dissociation Constant

• In dilute solutions the [H2O] is almost constant

• The expression can be rearranged so both constants are on the same side.

• The rearrangement gives Ka, the acid dissociation constant

[H2O]Kc = Ka = [H3O+][A-]

[HA]

Acid Dissociation Constant

• If you know [acid] and pH, you can find Ka

• A table of Ka values is located on Pg. 803 in your text.

Calculations with the Acid Dissociation Constant

• The smaller Ka is, the less the acid ionizes in aqueous solution

• Solving Equilibrium Problems Involving Acids and Bases

1.Write the balanced chemical equation2.Use the equation to write an ICE table3.Let x represent the change in concentration of

the substance with the smallest coefficient

• Solving Equilibrium Problems Involving Acids and Bases

4.If problem gives [inital] of the acid, compare [HA] with Ka

5.If [HA]/Ka > 100, the change in the [initial], x, is negligible and can be ignored.

6.If [HA]/Ka < 100 the change in [initial] may not be negligible. The equilibrium equation will be more complex and may require the solution of a quadratic equation.

Percent Dissociation

% Ionization = [molecules that ionize]x100%[Initial] Acid

Example Problem

• Propanoic Acid (CH3CH2COOH) is a weak monoprotic acid. A 0.10 mol/L solution has a pH of 2.96. What is Ka? What is the percent dissociation?

Given: Initial [CH3CH2COOH] = 0.10 mol/L pH = 2.96

Write the balanced equation.CH3CH2COOH(aq) + H2O(l) ↔ CH3CH2COO-

(aq) + H3O+ (aq)

• Prepare an ICE table

• Write the equation for KaKa = [CH3CH2COO-] [H3O+] = (x) (x)__ = x2___

[CH3CH2COOH] (0.10 – x) (0.10 – x)

• x = [H3O+] at equilibrium = 10-2.96 = 1.1x10-3mol/L

Concentration (mol/L)

CH3CH2COOH(aq) H2O(l) CH3CH2COO-(aq) H3O+

(aq)

Initial 0.10 0 ~0

Change -x +x +x

Equilibrium 0.10 - x +x +x

• Substitute value for x into the Ka expression.

Ka = _ x2___ = (1.1 x 10-3)2 = 1.2 x 10-5

(0.10 – x) (0.10 – 1.1 x 10-3)

Percent ionization = 1.1 x 10-3 mol/L x 100 0.10 mol/L = 1.1%

Therefore, Ka for propanoic acid is 1.2 x 10-5 and the percent ionization is 1.1%

Practice Problems

• Pg. 556 # 3, 5• Pg. 568 # 7, 8

Polyprotic Acids

• To calculate Ka,– Divide the problem into stages. – Equilibrium [acid] for the first H+ = initial [acid] for

the second H+

• To calculate [H3O+] and pH– With the exception of sulfuric acid, all polyprotic acids

are weak. The second dissociation is even weaker than the first

– Therefore, only the [first dissociation] is used to find [H3O+] and pH

Practice Problems

• Pg. 578 # 14Section Review:• Pg. 579 # 3, 4, 6, 13