mcdonald (pam78654) – HW 9A: Acids and Bases – laude – (89560) 1 This print-out should have 14 questions. 003 10.0 points Multiple-choice questions may continue on Which of the acids − the next column or page – find all choices I. HBrO3 II. GaCl3 III. HSO4 before answering. IV. AlF3 are classified as Lewis acids but are not 001 10.0 points Bronsted-Lowry acids? Which Ka value(s) would you use when calcu- lating the pH of a system involving Li2HPO4 1. I and II and Li3PO4? 2. II and IV correct 1. Ka2, Ka3 3. None of these 2. Ka1 4. II and III 3. Ka2 5. I and IV 4. K 3 correct a Explanation: Lewis acids accept a share in an electron 5. Ka1, Ka2 pair. A Bronsted-Lowry acid is a proton Explanation: donor. I and III would be Bronsted acids. Therefore, only II and IV can be the correct 002 10.0 points answer. One difference between a Lewis base and an Arrhenius base is 004 10.0 points Listed in order of increasing acid strength, 1. a Lewis base is a proton acceptor and which pair is incorrect? an Arrhenius base produces hydroxide ions in solution. 1. HNO3 < HNO2 correct 2. a Lewis base is an electron pair acceptor 2. H3AsO4 < H3PO4 and an Arrhenius base is a proton acceptor. 3. HF < HCl 3. a Lewis base is an electron pair donor and an Arrhenius base is a proton acceptor. 4. H3As < H2Se 4. a Lewis base is an electron pair acceptor 5. HClO < HClO2 and an Arrhenius base is a proton donor. Explanation: 5. a Lewis base is an electron pair donor and In a group of the Periodic Table, acid an Arrhenius base produces hydroxide ions in strength decreases as the metallic character of solution. correct element E in HxEOy increases. Acid strength increases as the number of oxygen atoms in Explanation: the formula increases for a given element X in A base is defined by the Arrhenius theory HXOy. HF is a weak acid and HCl is a strong as a substance which in water produces OH acid. ions, and by the Lewis theory as a species which provides an electron pair for sharing in 005 10.0 points a coordinate covalent bond. Which one of the following pairs of acids and their conjugate bases is INCORRECTLY mcdonald (pam78654) – HW 9A: Acids and Bases – laude – (89560) 2 matched? 1. Cannot be determined − 1. H2O:OH 2. II, I, III, IV + 2. H3O : H2O 3. None of these − 3. HClO : ClO 4. III, I, IV, II + − 4. NH4 : NH2 correct 5. II, IV, I, III − 5. HF:F Explanation: 6. IV, I, III, II An acid is a proton donor and a base is a 7. proton acceptor. The only difference between II, III, IV, I the acid and its conjugate base is that the − 8. IV, III, I, II correct base has one less H atom. NH2 has two fewer H atoms than the NH+; therefore, it is 4 9. incorrectly matched. I, IV, III, II 10. 006 10.0 points II, III, I, IV A0.0001 M solution of HCl has a pH of Explanation: The stronger the acid, the higher the Ka 1. 11. value and the lower the pKa value: K − K 2. 10. p a = log( a) −pKa Ka = 10 3. 3. I. For phosphorous acid, 4. 4. correct −2.00 Ka = 10 =0.01 Explanation: [HCl] = 0.0001 M II. For the hydrogen selenate ion, HCl is a strong acid which means it disso- + − −1.92 ciates completely into [H ] and [Cl ]. There- Ka = 10 =0.0120226 fore, we know that the concentration of [H+] is 0.0001 M. III. For phosphoric acid, −2.12 pH = − log[H+] = − log(0.0001) = 4 . Ka = 10 =0.00758578 IV. For selenous acid, 007 10.0 points −2.46 Ka = 10 =0.00346737 Arrange the acids I) phosphorous acid (H3PO3), − pKa1 =2.00; < < < − H2SeO3 H3PO4 H3PO3 HSeO4 II) hydrogen selenate ion (HSeO4 ), pKa =1.92; III) phosphoric acid (H3PO4), pKa =2.12; 008 10.0 points IV) selenous acid (H2SeO3), pKa =2.46; Which of in increasing order of strengths. I)HCl II)HF III)LiOH mcdonald (pam78654) – HW 9A: Acids and Bases – laude – (89560) 3 IV) HClO2 V) HNO3 corresponds to a strong acid. So really what are strong acids or strong bases in water? we’re looking for is which acid is strongest (has the lowest pH). 1. All of the compounds A low pH means that the [H+] concentra- tion is low. (Remember that values greater 2. I, III, and V only correct than 7 are basic!) The larger values of Ka means that there is more [H+] so you would 3. I, II, IV, and V only expect these solutions to be more acidic; i.e., have smaller pH’s. The smaller Ka values 4. I, III, IV, and V only mean less [H+] in solution, so higher pH’s. The acid with the largest Ka (#1) will have 5. I, II, III, and V only the lowest pH; i.e., highest [H+] concentration Explanation: 010 10.0 points × −9 009 10.0 points What is the pH of 2 10 M Ba(OH)2? Assume that five weak acids, identified only 1. 8.40 by numbers (1, 2, 3, 4 and 5), have the follow- ing ionization constants. 2. 8.70 Ionization 3. 7.02 correct Acid Constant Ka value 4. 5.60 − 1 1.0 × 10 3 − 2 3.0 × 10 5 5. 5.30 − 3 2.6 × 10 7 Explanation: −9 −9 −14 4 4.0 × 10 [Ba(OH)2]=2 × 10 M Kw =1 × 10 −11 5 7.3 × 10 Ba(OH)2 completely dissociates: 2+ − The anion of which acid is the weakest base? Ba(OH)2(aq) → Ba (aq)+ 2OH (aq) − 1. 3 [OH ] = 2[Ba(OH)2] − − = 2(2 × 10 9)=4 × 10 9 2. − 5 which is less than the [OH ] in pure water − (1 × 10 7), so we must consider this concen- 3. 2 tration: 4. 4 ⇀ + − H2O ↽ H + OH 5. 1 correct −7 −7 ini − 1 × 10 1 × 10 − − Explanation: ∆ − −4 × 10 9 +4 × 10 9 −8 −7 − fin 9.6 × 10 1.04 × 10 HA ↽⇀ H+ +A [H+][A−] Ka = − + [H] [A] Kw = [OH ][H ] The ‘anion of an acid’ is another way of saying Kw [H+] = ‘conjugate base,’ and a weak conjugate base [OH−] mcdonald (pam78654) – HW 9A: Acids and Bases – laude – (89560) 4 1 × 10−14 NaOH. Which of the following would be the = 1.04 × 10−7 same for both titrations? − =9.61538 × 10 8 1. the volume of NaOH added to reach the Thus equivalence point correct pH = − log[H+] 2. the pH at the halfway point − = − log(9.61538 × 10 8)=7.01703 3. the pH at the equivalence point Remember to check if your pH makes sense. The usual error here is to just substitute the 4. the initial pH − [OH ] from the Ba(OH)2 alone into the Kw expression to find [H+]; this gives an acidic 5. Two of the other answers are correct. pH value for the base Ba(OH)2. Explanation: Both HCl and CH2C3O2H2 are monoprotic 011 10.0 points acids and so if the volume concentrations are + For a solution labeled “0.10 M H2SO4(aq),” identical, the amounts of replaceable H in both solutions are identical. − 1. [HSO4 ] is greater than 0.10 M. 014 10.0 points 2. the pH is less than 1.0. correct What would be the pH of a solution of hypo- bromous acid (HOBr) prepared by dissolving 2− 3. [SO4 ] = 0.10 M. 9.7 grams of the acid in 20 mL of pure wa- ter (H2O)? The Ka of hypobromous acid is − 4. the pH equals 1.0. 2 × 10 9 5. the pH is greater than 1.0. 1. 13 Explanation: 2. 1 012 10.0 points 3. 10 Estimate the pH of 0.10 M Na2HPO4(aq) given pK =2.12, pK =7.21, and pK = a1 a2 a3 4. 4 correct 12.68 for phosphoric acid. 5. 6 1. 4.67 Explanation: 1 mol 2. 7.40 9.7 g HOBr × =0.1 mol HOBr 97 g 3. 9.94 correct 0.1 mol HOBr = 5 M HOBr 0.02LH2O 4. 2.12 + 1/2 −9 1/2 [H ]=(Ka · Ca) = (2 × 10 · 5) 5. 12.68 − − = (10 8)1/2 = 10 4 Explanation: − pH = − log[H+] = − log(10 4)=4 013 10.0 points Consider the titration of equal volumes of 0.1 M HCl and 0.1 M HC2H3O2 with 0.1 M.