Ch 12.7 Acids and Bases Supplemental Instruction Iowa State University Leader: Course: Instructor: Date: Grant DeRocher Chem 167 Houk 04/21/13 Know Kw= 1.0 x 10-14 12.71, 12.73, 12.74, 12.75, 12.79, 12.80, 1. Write the formula of the conjugate base of each of the following acids: (a)HNO3 (b)H2O (c) HSO4- (d) H2CO3 (e) H3O+ (a) NO3(b) OH(c) SO42(d) HCO3(e) H2O 2. For each of the following reactions, indicate the Bronsted-Lowry acids and bases. What are the conjugate acid-base pairs? (a) CN-(aq) + H2O(l) HCN(aq) + OH-(aq) (b) HCO3-(aq) + H3O+(aq) H2CO3-(aq) + H2O(l) (c) CH3COOH(aq) + HS-(aq) CH3COO-(aq) + H2S(aq) (a) CN-(aq) BL BASE+ H2O(l)BL ACID HCN(aq) conj acid+ OH-(aq) conj base (b) HCO3-(aq) + H3O+(aq) H2CO3-(aq) + H2O(l) (c) CH3COOH(aq) + HS (aq) CH3COO-(aq) + H2S(aq) Other two work the same as the first one. Use bronsted Lawry definition of acids and bases 3. What are the products of the following acid-base reactions? Indicate the acid and its conjugate base and the base and its conjugate acid. (a) HClO4 + H2O (b) NH4+ + H2O (c) HCO3- + OH(a) ClO4- + H3O+ (b) NH3 + H3O+ (c) CO32- + H2O 4. Write chemical equations and equilibrium expressions for the reactions of each of the following weak acids with water: (a) CH3COOH (b) C2H5COOH (c) HF (d) HClO (e) H2CO3 Ka= [products]/[reactants] make sure to write the equation first to find Ka H2O becomes H3O+(aq) 5. Hydrofluoric acid is a weak acid used in the building industry to etch patterns into glass for elegant windows. Because it dissolves glass, it is the only inorganic acid that must be stored in plastic containers. A 0.1 M solution of HF has a pH of 2.1. Calculate [H3O+] in this solution. The definition of pH is: pH=-log[H3O+] rearranging [H3O+] = 10-pH pH=2.1 [H3O+] = 10-2.1 = 8x10-3 (one sig fig) 6. Calculate the pH of a 0.10 M solution of propanoic acid and determine its present ionization. The ionization constant Ka for propanoic acid is 1.3x10-5. C2H5COOH(aq) + H2O C2H5COO-(aq) + H3O+ Initial 0.10 M 0.00 M 0.00 M Change -X +X +X Equilibrium 0.10 M X X The equilibrium constant expression is: Ka= [C2H5COO-][H3O+]/[C2H5COOH]=1.3x10-5 [X][X]/[0.10-X]=1.3x10-5 Assume that .10-X is approximately .10. Ka=X2/.10=1.3x10-5. X2=1.3x10-6 so X=1.1x10-3 M = [H3O+] = [C2H5COO-] The pH= -log[H3O+]= -log[1.1x10-3]= 2.9 Percent ionization is defined as % ionization = [Anion]eq/[acid]init x 100 [C2H5COO-]/[ C2H5COOH] x 100 = [1.1x10-3/[ .10] x 100= 1.1% 7. Acrylic acid is used in the polymer industry in the production of acrylates. Its Ka is 5.6x10-5 . What is the pH of a 0.11 M solution of acrylic acid, CH2CHCOOH? CH2CHCOOH(aq) + H2O Initial 0.11 M Change -X Equilibrium 0.11 M CH2CHCOO-(aq) + H3O+ 0.00 M 0.00 M +X +X X X The equilibrium constant expression is: Ka= [CH2CHCOO-][H3O+]/[ CH2CHCOOH]=5.6x10-5 [X][X]/[0.10-X]=5.6x10-5 Assume that .10-X is approximately .10. Ka=X2/.10=5.6x10-5. X2=6.2x10-6 so X=2.5x10-3 M = [H3O+] = [CH2CHCOOH] The pH= -log[H3O+]= -log[2.5x10-3]= 2.6 8. Morphine, an opiate derived from opium poppy, has the molecular formula C7H19NO3. It is a weakly basic amine, with a Kb of 1.6x10-6. What is the pH of a 0.0045 M solution of morphine? C2H19NO3 Initial Change Equilibrium + C7H20NO3+ 0.00 M +X X H2O 0.0045 M -X 0.0045 M + OH0.00 M +X X The equilibrium constant expression is: Ka= [C7H20NO3+][OH-]/[ C2H19NO3]=1.6x10-6 [X][X]/[0.10-X]=1.6x10-6 Assume that .10-X is approximately .10. Ka=X2/.10=1.6x10-6. X2=7.2x10-9 so X=8.5x10-5 M = [OH-] = [C7H20NO3+] We need to calculate the [H3O+] to find the pH. [H3O+][OH-] =Kw = 1.0x10-14 , [H3O+]=Kw/[OH-]= 1.0x10-14 / 8.5x10-5= 1.2x10-10 M The pH= -log[H3O+ ]= -log[1.2x10-10]= 9.9