Presentation on theme: "8.2 – Equilibrium of Weak Acids and Bases"— Presentation transcript:
18.2 – 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.
2Dissociation of WaterPure water contains a few ions from the dissociation of water.2H20(l) ↔ H3O+(aq) OH-(aq)At 25°C ~ 2 molecules/1 billion dissociateDue to the 1:1 ratio of H3O+ to OH- the [H3O+] = [OH-]At 25°C, [H3O+] = [OH-] = 1.0x10-7 mol/L
3The Ion Product Constant for Water The equilibrium constant for waterKc = [H3O+] [OH-][H2O]2The 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 (units are dropped)
4Strong Acids and BasesWith 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/LSo it can be ignored
5[H3O+] and [OH-] at 25°C Acidic Solutions Neutral Solutions [H3O+] > 1.0 x 10-7 mol/L[OH-] < 1.0 x 10-7 mol/LNeutral Solutions[OH-] = [H3O+] = 1.0 x 10-7 mol/LBasic Solutions[H3O+] < 1.0 x 10-7 mol/L[OH-] > 1.0 x 10-7 mol/L
6Determining [H3O+] and [OH-] Find [H3O+] and [OH-] in 0.16 M Ba(OH)2Ba(OH)2 is a strong base, therefore it dissociates completely.Therefore use [Ba(OH)2] to find [OH-]H2OBa(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
7Determining [H3O+] and [OH-] Use Kw = [H3O+] [OH-] = 1.0 x to find [H3O+][H3O+][OH-] = 1.0 x 10-14[H3O+] = 1.0 x 10-14[OH-][H3O+] = 1.0 x = 3.1 X mol/L0.32 mol/LTherefore the [H3O+] = 3.1 x mol/L, and the[OH-] = 0.32 mol/L
8Practice Finding [OH-] and [H3O+] for strong acids and bases Practice problems on Pg. 537 # 4, 5Pg. 540 # 10
9Calculating pH and pOH This should be review from SCH3U pH = -log[H3O-]pOH = -log[OH-]Kw = [H3O+] [OH-] = 1.0 x at 25°CTherefore, pH + pOH = 14
10Problems involving pH and pOH A liquid shampoo has a hydroxide ion concentration of 6.8 x 10-5 mol/L at 25°CIs the shampoo acidic, basic or neutral[OH-] = 6.8 x 10-5 mol/L > 1.0 x 10-7 mol/LTherefore, the shampoo is basicCalculate the hydronium ion concentration.[H3O+] = 1.0 x = 1.5 x mol/L6.8 x 10-5
11Problems involving pH and pOH A liquid shampoo has a hydroxide ion concentration of 6.8 x 10-5 mol/L at 25°CWhat is the pH and the pOH of the shampoo?pH = -log[H3O+] = -log[1.5 x 10-10] = 9.83pOH = -log[OH-] = -log[6.8 x 10-5] = 4.17Note: With pH and pOH values. The numbers to the left of the decimal do not count as sig. digits.
12Alternative Method for finding [H3O+] and [OH-] [H3O+] = 10-pH[OH-] = 10-pOHEx. If the pH is 5.20, what is the [H3O+] = 10-pH[H3O+] = = 6.3 x 10-6 mol/LPractice ProblemsPg. 546 # 12, 13Pg. 549 # 17, 18
13Acid 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 isKc = [H3O+][A-][HA][H2O]
14Acid Dissociation Constant In dilute solutions the [H2O] is almost constantThe 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]
15Acid Dissociation Constant If you know [acid] and pH, you can find KaA table of Ka values is located on Pg. 803 in your text.
16Calculations with the Acid Dissociation Constant The smaller Ka is, the less the acid ionizes in aqueous solutionSolving Equilibrium Problems Involving Acids and BasesWrite the balanced chemical equationUse the equation to write an ICE tableLet x represent the change in concentration of the substance with the smallest coefficient
17Solving Equilibrium Problems Involving Acids and Bases If problem gives [inital] of the acid, compare [HA] with KaIf [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.
18Percent Dissociation% Ionization = [molecules that ionize]x100% [Initial] Acid
19Example ProblemPropanoic Acid (CH3CH2COOH) is a weak monoprotic acid. A 0.10 mol/L solution has a pH of What is Ka? What is the percent dissociation?Given: Initial [CH3CH2COOH] = 0.10 mol/LpH = 2.96Write the balanced equation.CH3CH2COOH(aq) + H2O(l) ↔ CH3CH2COO-(aq) + H3O+ (aq)
20Write the equation for Ka Ka = [CH3CH2COO-] [H3O+] = (x) (x)__ = x2___ Prepare an ICE tableWrite the equation for KaKa = [CH3CH2COO-] [H3O+] = (x) (x)__ = x2___[CH3CH2COOH] (0.10 – x) (0.10 – x)x = [H3O+] at equilibrium = = 1.1x10-3mol/LConcentration (mol/L)CH3CH2COOH(aq)H2O(l)CH3CH2COO-(aq)H3O+ (aq)Initial0.10~0Change-x+xEquilibriumx
21Substitute value for x into the Ka expression. Ka = _ x2___ = (1.1 x 10-3) = 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 1000.10 mol/L= 1.1%Therefore, Ka for propanoic acid is 1.2 x 10-5 and the percent ionization is 1.1%
23Polyprotic Acids To calculate Ka, To calculate [H3O+] and pH Divide the problem into stages.Equilibrium [acid] for the first H+ = initial [acid] for the second H+To calculate [H3O+] and pHWith the exception of sulfuric acid, all polyprotic acids are weak. The second dissociation is even weaker than the firstTherefore, only the [first dissociation] is used to find [H3O+] and pH