1 / 24

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.

marvin-richardson
Télécharger la présentation

8.2 – Equilibrium of Weak Acids and Bases

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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

More Related