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8.2 – Equilibrium of Weak Acids and Bases

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.

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8.2 – Equilibrium of Weak Acids and Bases

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  1. 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.

  2. 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

  3. 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)

  4. 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

  5. [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

  6. 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

  7. 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

  8. Practice • Finding [OH-] and [H3O+] for strong acids and bases • Practice problems on Pg. 537 # 4, 5 • Pg. 540 # 10

  9. 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

  10. Problems involving pH and pOH • A liquid shampoo has a hydroxide ion concentration of 6.8 x 10-5 mol/L at 25°C • Is the shampoo acidic, basic or neutral [OH-] = 6.8 x 10-5 mol/L > 1.0 x 10-7 mol/L Therefore, the shampoo is basic • Calculate the hydronium ion concentration. [H3O+] = 1.0 x 10-14 = 1.5 x 10-10 mol/L 6.8 x 10-5

  11. Problems involving pH and pOH • A liquid shampoo has a hydroxide ion concentration of 6.8 x 10-5 mol/L at 25°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.

  12. 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/L Practice Problems • Pg. 546 # 12, 13 • Pg. 549 # 17, 18

  13. Acid Dissociation Constant • Weak acids do not completely dissociate in water. • For a weak monoprotic acid HA(aq) + H2O(aq) ↔ H3O+(aq) + A-(aq) • The equilibrium expression is Kc = [H3O+][A-] [HA][H2O]

  14. 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]

  15. 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.

  16. 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 • Write the balanced chemical equation • Use the equation to write an ICE table • Let x represent the change in concentration of the substance with the smallest coefficient

  17. Solving Equilibrium Problems Involving Acids and Bases • If problem gives [inital] of the acid, compare [HA] with Ka • If [HA]/Ka > 100, the change in the [initial], x, is negligible and can be ignored. • 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.

  18. Percent Dissociation % Ionization = [molecules that ionize]x100% [Initial] Acid

  19. 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)

  20. Prepare an ICE table • Write the equation for Ka Ka = [CH3CH2COO-] [H3O+] = (x) (x)__ = x2___ [CH3CH2COOH] (0.10 – x) (0.10 – x) • x = [H3O+] at equilibrium = 10-2.96 = 1.1x10-3mol/L

  21. 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%

  22. Practice Problems • Pg. 556 # 3, 5 • Pg. 568 # 7, 8

  23. 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

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

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