FORMULA SHEET (tear off) N A = 6.022 x 10 23 C = ( 5 / 9) ( F - 32) F = ( 9 / 5)( C) + 32 1 amu = 1.661 x 10-27 kg C = K - 273.15 K = C + 273.15 1 atm = 760 torr = 760 mm Hg 1 atm = 1.013 bar pv = nrt R = 0.08206 L atm/mol K 1 L atm = 101.3 J R = 8.314 J/mol K 1 J= 1 kg m 2 /s 2 ln(p) = - H vap + C ln(p 2/p 1) = - ( H vap/r) { (1/T 2) - (1/T 1) } T p A = X A p A [B] = k p B p A = X Bp A T b = K b m B T f = K f m B = [B]RT H = U + pv G = H - TS G rxn = G rxn + RT ln Q ln K = - G rxn/rt K p = K C (RT) n If ax 2 + bx + c = 0, then x = ( - b [b 2-4ac] 1/2 ) 2a K a K b = K w = 1.0 x 10-14 (at T = 25 C) ph = pk a + log 10{[base]/[acid]}
GENERAL CHEMISTRY 2 SECOND HOUR EXAM June 8, 2018 Name Version 1 Panthersoft ID Signature Part 1 (24 points) Part 2 (44 points) Part 3 (52 points) TOTAL (120 points) Do all of the following problems. Show your work. Unless otherwise specified, you may assume T = 25.0 C in all of the problems on this exam. 2
Part 1. Multiple choice. Circle the letter corresponding to the correct answer. There is one and only one correct answer per problem. [4 points each] 1) An Arrhenius acid is best defined as a) A substance that produces H + ions when dissolved in water b) A substance that produces OH - ions when dissolved in water A c) A proton donor d) A proton acceptor e) An electron pair donor ----------------------------------------------------------------------------------------------------------------------------------------- The following information will be of use in doing problems 2 and 3 below. Weak acid K a Weak acid K a benzoic acid (C 6H 5COOH) 6.5 x 10-5 nitrous acid (HNO 2) 4.5 x 10-4 chloroacetic acid (CH 2ClCOOH) 1.4 x 10-3 phenol (C 6H 5OH) 1.3 x 10-10 ----------------------------------------------------------------------------------------------------------------------------------------- 2) Which of the following 0.0100 M solutions of weak acid will have the lowest value for ph? a) a 0.0100 M solution of benzoic acid b) a 0.0100 M solution of chloroacetic acid B c) a 0.0100 M solution of nitrous acid d) a 0.0100 M solution of phenol e) All of the above solutions will have the same value for ph 3) Which of the following 0.0100 M solutions of the sodium salt of the conjugate base of a weak acid will have the highest value for ph? a) a 0.0100 M solution of sodium benzoate (NaC 6H 5COO) b) a 0.0100 M solution of sodium chloroacetate (NaCH 2ClCOO) D c) a 0.0100 M solution of sodium nitrite (NaNO 2) d) a 0.0100 M solution of sodium phenolate (NaC 6H 5O) e) All of the above solutions will have the same value for ph 4) Which of the following metal ions would be expected to be the strongest weak acid? a) Al 3+ ion b) Ba 2+ ion A c) Cs + ion d) Cu 2+ ion e) Pb 2+ ion 5) Which of the following reactions is expected to go to completion? a) The reaction of a strong acid with a strong base b) The reaction of a strong acid with a weak base E c) The reaction of a weak acid with a strong base d) Both a and b e) Both a and b and c 6) For a supersaturated solution a) Q sp < K sp b) Q sp = K sp C c) Q sp > K sp d) All of the above can be true e) None of the above can be true Version 2: B C D E E A Version 3: C A C B E A Version 4: D A B D E B 3
Part 2. Short answer questions. 1) An aqueous solution has a ph = 8.17. Give the poh and the concentrations of H 3O + and OH - ions in the solution. [8 points] poh = 5.83 [H 3O + ] = 6.8 x 10-9 [OH - ] = 1.5 x 10-6 Version 2: 6.14, 1.4 x 10-8, 7.2 x 10-7 Version 3: 6.38, 2.4 x 10-8, 4.2 x 10-7 Version 4: 5.52, 3.3 x 10-9, 3.0 x 10-6 2) Give the following (including correct charge) [3 points each] a) The conjugate acid of the hydrogen phosphate ion, HPO 3 2-. H2PO3 - b) The conjugate base of bromous acid, HBrO 2, BrO2-3) A solution is prepared by dissolving 2.186 g of barium hydroxide (Ba(OH) 2, MW = 171.3 g/mol), a strong soluble base, in water. The final volume of the solution formed is V = 400.0 ml. What is the value of ph for the solution? [8 points] moles OH - = 2.186 g Ba(OH) 2 1 mol 2 mol OH - = 0.0255 mol 171.3 g 1 mol Ba(OH) 2 [OH - ] = 0.0255 mol OH - = 0.0638 M poh = - log 10(0.0638) = 1.20 0.4000L ph = 14.00-1.20 = 12.80 Version 2: 12.77 Version 3: 12.79 Version 4: 12.83 4) For each of the following circle the correct answer. There is one and only one correct answer per problem. [3 points each] a) The acid that is a strong acid (would also accept HBrO3 as correct) HBrO 3 HF HNO3 H 2SeO 4 b) The ionic compound that is a strong soluble base AgOH Co(OH) 2 Fe(OH) 3 Sr(OH)2 c) The substance where phosphorus (P) has the largest value for oxidation number P2O5 H 3PO 3 PF 3 P 4 d) The strongest weak acid HBrO 2 HBrO3 HIO 2 HIO 3 4
5) The solubility product for calcium fluoride (CaF 2, MW = 78.1 g/mol) is K sp = 4.2 x 10-11. How many grams of calcium fluoride can dissolve in 1.000 L of pure water? [10 points] CaF 2(s) Ca 2+ (aq) + 2 F - (aq) Ksp = [Ca 2+ ] [F-] 2 = 4.2 x 10-11 Initial Change Equilibrium Ca 2+ 0 x x So (x) (2x) 2 = 4x 3 = 4.2 x 10-11 F - 0 2x 2x x 3 = 4.2 x 10-11 = 1.05 x 10-11 x = (1.05 x 10-11 ) 1/3 = 2.19 x 10-4 But x = [Ca 2+ ] = moles of CaF 2 that dissolves per liter of solution, and so mass CaF 2 = 1.000 L soln 2.19 x 10-4 mol 78.1 g = 1.71 x 10-2 g L soln mol CaF 2 Part 3. Problems 1) A 25.00 ml sample of a stock solution of potassium hydroxide (KOH, MW = 56.11 g/mol) is titrated with a 0.2281 M solution of hydrochloric acid (HCl, MW = 36.46 g/mol). After the addition of 23.04 ml of the HCl solution the equivalence point for the titration is reached. What is the concentration of the stock solution of potassium hydroxide? [12 points] reaction is HCl + KOH H 2O + KCl moles HCl = 0.02304 L soln 0.2281 mol = 5.255 x 10-3 mol HCl L soln From the balanced equation, moles KOH = moles HCl = 5.255 x 10-3 mol So [KOH] = 5.255 x 10-3 mol KOH = 0.2102 M 0.02500 L soln Version 2: 0.2976 M Version 3: 0.1701 M Version 4: 0.1870 M 5
2) Propanoic acid (CH 3CH 2COOH, MW = 74.08 g/mol) is a weak monoprotic acid, with K a = 1.34 x 10-5. a) A chemist prepares 1.000 L of a 0.0360 M aqueous solution of propanoic acid. What is the ph of the solution? [12 points] CH 3CH 2COOH(aq) + H 2O( ) H 3O + (aq) CH 3CH 2COO - (aq) K a = [H 3O + ] [CH 3CH 2COO - ] = 1.34 x 10-5 [CH 3CH 2COOH] Initial Change Equilibrium H 3O + 0 x x So (x) (x) 1.34 x 10-5 CH 3CH 2COO - 0 x x (0.0360 - x) CH 3CH 2COOH 0.0360 - x 0.0360 - x Assume x << 0.0360 Then x 2 = 1.34 x 10-5 x 2 = (0.0360)(1.34 x 10-5 ) = 4.82 x 10-7 0.0360 x = (4.82 x 10-7 ) 1/2 = 6.95 x 10-4 The assumption that x was small was good. [H 3O + ] = x = 6.95 x 10-4 ph = - log 10(6.95 x 10-4 ) = 3.16 Version 2: 3.03 Version 3: 3.10 Version 4: 2.98 b) 0.500 g of sodium hydroxide (NaOH, MW = 40.00 g/mol) is added to the above 1.000 L solution of propanoic acid. What is the new value for ph after the addition of the sodium hydroxide? You may assume that the addition of NaOH does not change the volume of the solution. [12 points] The neutralization reaction that occurs is CH 3CH 2COOH(aq) + NaOH CH 3CH 2COO - (aq) + H 2O(l) initial moles NaOH = 0.500 g 1 mol = 0.0125 mol NaOH 40.00 g initial moles CH 3CH 2COOH = 1.000 L soln 0.0360 mol = 0.0360 moles CH 3CH 2COOH L soln After neutralization, the moles of CH 3CH 2COOH is 0.0360 moles - 0.0125 moles = 0.0235 moles The reaction produces conjugate base, so after the neutralization we have 0.0125 moles CH 3CH 2COO - We can use the Henderson equation to find the ph ph = pk a + log 10{base/acid} = - log 10(1.34 x 10-5 ) + log 10{0.0125/0.0235} = 4.87 + (- 0.27) = 4.60 Version 2: 4.68 Version 3: 4.42 Version 4: 4.51 6
3) Consider the following unbalanced oxidation-reduction reaction NO 3 - (aq) + Zn(s) NH 4 + (aq) + Zn 2+ (aq) a) The oxidizing agent in the above reaction is (circle the correct answer) [4 points] NO3 - Zn NH 4 + Zn 2+ b) Balance the above unbalanced equation for acid conditions [12 points] oxidation Zn(s) Zn 2+ (aq) + 2 e - x 4 reduction net NO 3 - (aq) + 10 H + (aq) + 8 e - NH 4 + (aq) + 3 H 2O(l) 4 Zn(s) + NO 3 - (aq) + 10 H + (aq) 4 Zn 2+ (aq) + NH 4 + (aq) + 3 H 2O(l) 7