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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
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