FRONT PAGE FORMULA SHEET - TEAR OFF

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1 FRONT PAGE FORMULA SHEET - TEAR OFF N A = x C = ( 5 / 9) ( F - 32) F = ( 9 / 5)( C) amu = x kg C = K K = C atm = 760 torr = 760 mm Hg 1 atm = bar pv = nrt R = L. atm/mol. K 1 L. atm = J 1 amp = 1 C/s R = J/mol. K 1 J= 1 kg. m 2 /s 2 (1 volt) (1 Coulomb) = 1 Joule p A = X A p A [B] = K B p B p A = X Bp A T b = K b m B T f = K f m B = [B]RT [A] t = [A] 0 e -kt ln[a] t = ln[a] 0 - kt t 1/2 = ln2/k [A] t = [A] 0/(1 + kt[a] 0) (1/[A] t) = (1/[A] 0) + kt t 1/2 = 1/(k[A] 0) k = A e -Ea/RT ln k = ln A - (E a/r)(1/t) ln(k 2/k 1) = - (E a/r) [ (1/T 2) - (1/T 1) ] If ax 2 + bx + c = 0, then x = ( - b [b 2-4ac] 1/2 )/2a K P = K c (RT) n K a. K b = K w = 1.0 x (at T = 25 C) ph = pk a + log 10{[base]/[acid]} H = E + pv G = H - TS F = C/mol G rxn = G rxn + RT ln Q ln K = - G rxn /RT G = - nfe E = E - (RT/nF) ln Q ln K = nfe /RT

2 GENERAL CHEMISTRY 2 SECOND EXAM Name Panthersoft ID Signature Part 1 (xx points) Part 2 (xx points) Part 3 (xx points) TOTAL (xxx points) Do all of the following problems. Show your work. 2

3 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) Which of the following is classified as a strong acid a) HNO 2 b) HClO 2 C c) HCl d) Both b and c e) Both a and b and c 2) In the reaction Ag + (aq) + 2 NH 3(aq) Ag(NH 3) 2 + (aq) Ag + (aq) is a) a Bronsted base b) a Bronsted acid D c) a Lewis base d) a Lewis acid e) None of the above 3) A L sample of a buffer solution contains moles of acetic acid (CH 3COOH), a weak acid, and moles of sodium acetate (NaCH 3COO) a soluble ionic compound ml of a 1.00 M solution of sodium hydroxide (NaOH), a strong soluble base, is added to the solution. The ph of the solution a) will increase by a large amount (more than 0.50 ph units). b) will increase by a small amount (less than 0.10 ph units). B c) will decrease by a large amount (more than 0.50 ph units). d) will decrease by a small amount (less than 0.10 ph units). e) will not change. 4) Potassium hydrogen phthalate (KHP, M = g/mol) is a weak monoprotic acid often used to find the concentrations of stock solutions of strong soluble bases. Consider the titration of a 1.5 g sample of KHP with a stock solution of NaOH, a strong soluble base, whose concentration is approximately 0.10 M. Which of the following would be the best indicator for the titration? a) Thymol blue (pk a = 2.0) b) Bromcresol blue (pk a = 4.5) D c) Bromthymol blue (pk a = 6.6) d) o-cresolphthalein (pk a = 9.0) e) All of the above indicators would be equally good for finding the equivalence point for this titration 5) Consider the reaction forming the complex ion Cu(CN) 4 2- (aq) Cu 2+ (aq) + 4 CN - (aq) Cu(CN) 4 2- (aq) K f = 1.0 x (at T = 25. C) A system initially contains M Cu 2+ ion and M CN - ion at T = 25. C. At equilibrium a) the concentration of Cu 2+ ion will be small (less than 10-4 M) b) the concentration of CN - ion will be small (less than 10-4 M) B c) the concentration of Cu(CN) 4 2- ion will be small (less than 10-4 M) d) Both a and b e) None of the ion concentrations will be small 3

4 6) For a particular process S system > 0. The value for S surr a) must be positive ( S surr > 0) b) must be exactly equal to zero E c) must be negative ( S surr < 0) d) can be either positive or zero e) can be either positive, zero, or negative 7) For which of the following substances will the free energy of formation ( G f) at T = 25. C be equal to zero? a) Ag(s) b) C 10H 8(s) A c) SO 2(g) d) Both a and b e) Both a and b and c 8) For a particular chemical reaction H rxn > 0 and S rxn > 0. Based on this we would expect that for standard conditions a) the reaction will always be spontaneous. b) the reaction will never be spontaneous. C c) the reaction will be spontaneous at high temperature but not at low temperature. d) the reaction will be spontaneous at low temperature but not at high temperature e) Cannot tell from the information given. Part 2. Short answer. 1) For each of the following circle the correct answer. There is one and only one correct answer per question. [3 points each] a) The strongest acid HBrO3 HBrO 2 HIO 3 HIO 2 b) The strongest acid H2Te H 2Se H 2S H 2O c) An example of a strong soluble base Co(OH) 2 Cu(OH) 2 Sr(OH)2 Zn(OH) 2 d) The aqueous solution with the lowest value for ph (at T = 25.0 ) M KClO M KI M NH 4ClO M NH4I 2) Give the chemical formula for the conjugate acid and conjugate base of C 6H 7O 6 (including correct charge). [4 points] conjugate acid C 6H 8O 6 conjugate base C 6H 6O 6 2-4

5 3) Define the following terms [4 points] a) end point The point in a titration where the indicator changes color. We pick an indicator so that the volume of titrant needed to reach end point and the equivalence point are as close as possible. b) Second Law of thermodynamics - S univ 0 for any allowed process. 4) Find the ph of a mol/l aqueous solution of potassium hydroxide, KOH, a strong soluble base, at T = 25.0 C. [6 points] KOH is a strong soluble base, and so reacts as follows KOH(s) K + (aq) + OH - (aq) So the concentration of OH - in the system is the same as the molarity of KOH. And so [OH - ] = poh = - log 10(0.0400) = 1.40 ph = poh = ) A mol/l aqueous solution of a weak monoprotic acid HA, has ph = 3.76 at T = 25.0 C. What is the numerical value for K a for the acid at this temperature? [8 points] The equilibrium is HA(aq) + H 2O(l) H 3O + (aq) + A - (aq) K a = [H 3O + ] [A - ] [HA] Initial Change Equilibrium H 3O + 0 x x A x HA x x We don t know K a, however, we do know ph. And so [H 3O + ] = x = 10 -ph = = 1.74 x 10-4 Substitutig for x in the expression for K a gives K a = (1.74 x 10-4 )(1.74 x 10-4 ) = 1.2 x 10-7 ( x 10-4 ) 5

6 6) A buffer contains significant amounts of hypobromous acid (HBrO), a weak acid, and sodium hypobromite (NaBrO). Write equations showing how this buffer neutralizes added strong acid and added strong base. [8 points] strong acid (example, HCl) HCl(aq) + BrO - (aq) HBrO(aq) + Cl - (aq) converts strong acid into weak acid strong base (example, NaOH) NaOH(aq) + HBrO(aq) Na + (aq) + OBr - (aq) + H 2O(l) converts strong base into weak base 7) The solubility of aluminum fluoride (AlF 3, MW = g/mole) is 5.59 g/l at T = 25. C. What is the value for K sp, the solubility product, for AlF 3? [8 points] AlF 3(s) Al 3+ (aq) + 3 F - (aq) K sp = [Al 3+ ][F - ] 3 Initial Change Equilibrium Al 3+ 0 x x F - 0 3x 3x (x)(3x) 3 = 27 x 4 = K sp But x is equal to the molar solubility, which is mol sol = 5.59 g 1 mol L g = mol/l K sp = 27 (0.0666) 4 = 5.3 x ) What is the oxidation number for sulfur (S) in each of the following molecules or ions? [3 points each] H 2S -2 S 2O H 2SO 3 +4 S 8 0 9) Balance the following oxidation-reduction reaction for the indicated condition. [12 points each] H 3AsO 3(aq) + Cr 2O 7 2- (aq) H 3AsO 4(aq) + Cr 3+ (aq) (in acid solution) 3 x ox H 3AsO 3(aq) + H 2O(l) H 3AsO 4(aq) + 2 e H + (aq) red Cr 2O 7 2- (aq) + 6 e H + (aq) 2 Cr 3+ (aq) + 7 H 2O(l) net 3 H 3AsO 3(aq) + Cr 2O 7 2- (aq) + 8 H + (aq) 3 H 3AsO 4(aq) + 2 Cr 3+ (aq) + 4 H 2O(l) 6

7 Part 3. Problems. 1) Pentanoic acid (C 4H 9COOH, M = g/mol) is a weak monoprotic acid used as the starting material in the synthesis of additives used in some perfumes. It has an acid ionization constant K a = 1.52 x 10-5 at T = 25.0 C. a) Write the balanced chemical equation showing how pentanoic acid functions as a weak acid according to the Bronsted definition of acids and bases. [3 points] C 4H 9COOH(aq) + H 2O(l) H 3O + (aq) + C 4H 9COO - (aq) K a = [H 3O + ][C 4H 9COO - ] [C 4H 9COOH] b) Find the ph and percent ionization for a mol/l aqueous solution of pentanoic acid at T = 25.0 C. [12 points] If we set up our ICE table, we get Initial Change Equilibrium H 3O + 0 x x C 4H 9COO - 0 x x C 4H 9COOH x x So (x)(x) = 1.52 x 10-5 ( x) If we assume x << , we get x 2 = (0.0360)(1.52 x 10-5 ) = 5.47 x 10-7 x = (5.47 x 10-7 ) 1/2 = 7.4 x 10-4 ph = - log 10(7.4 x 10-4 ) = 3.13 % ionization = { [C 4H 9COO] eq//[c 4H 9COOH] init } x 100. % = 2.1% 7

8 2) Nitrous acid (HNO 2, M = g/mol) is a weak monoprotic acid, with K a = 4.6 x 10-4 at T = 25. C. A student adds 4.04 g of potassium hydroxide (KOH, M = 56.1 g/mol), a strong soluble base, to ml of a M aqueous solution of nitrous acid. The addition of KOH causes no significant change in the volume of the solution. a) Write the reaction that occurs when the KOH is added to the nitrous acid solution. [4 points] HNO 2(aq) + KOH(aq) K + (aq) + NO 2- (aq) + H 2O(l) b) What is the ph of the above solution after the addition of KOH? [10 points] moles HNO 2 = ( mol/l) ( L) = mol moles KOH = 4.04 g (1 mol/56.1 g) = mol So KOH is the limiting reactant in the above reaction. We may make a table of the number of moles before and after neutralization. Initial After neutralization HNO KOH NO The solution is a buffer, so we may use the Henderson equation to find ph ph = pk a + log 10{[base]/[acid]} = { - log 10(4.6 x 10-4 ) + log 10{(0.0720)/(0.1414)} = 3.04 c) Does the addition of the potassium hydroxide lead to the formation of a buffer solution (yes/no, and a justification for your answer)? [4 points] Yes, since after addition of the KOH there are appreciable amounts of both a weak acid (HNO 2) and its conjugate base (NO 2 - ). 8

9 3) The thermochemical data given below is at T = 25. C, and may be used to do the following problem. Substance M(g/mol) H (kj/mol) G (kj/mol) S (J/mol. K) Ca 2+ (aq) CaCl 2(s) CaF 2(s) HCl(g) HF(g) Calcium fluoride (CaF 2) and calcium chloride (CaCl 2) are both ionic solids. In the presence of the gas phase compounds HF and HCl, the following equlibrium will be established. CaF 2(s) + 2 HCl(g) CaCl 2(s) + 2 HF(g) a) Give the expression for K, the thermodynamic equilibrium constant, for the above reaction. [4 points] K = (p HF) 2 (p HCl) 2 b) What is the numerical value for K for the above reaction at T = 25. C? [10 points] G rxn = [ ( G f(cacl 2(s)) + 2( G f(hf(g)) ] [ ( G f(caf 2(s)) + 2( G f(hcl(g)) ] = [ ( kj/mol) + 2( kj/mol) ] [ ( kj/mol) + 2( kj/mol) ] = 63.4 kj/mol lnk = - ( G rxn/rt) = J/mol = K = e = 7.7 x (8.314 J/mol. K)(298 K) 9

10 4) Consider the following galvanic cell (at T = 25. C) Cu(s) Cu 2+ ( M) Ag + ( M) Ag(s) Find the following. You may need to use some of the reduction data given below in doing part of this problem Ag + (aq) + e - Ag(s) Cu + (aq) + e - Cu(s) Cu 2+ (aq) + 2e - Cu(s) Cu 2+ (aq) + e - Cu + (aq) E = 0.80 v E = 0.52 v E = 0.34 v E = 0.16 v a) The half cell oxidation reaction, the half cell reduction, and the net cell reaction [5 points] ox Cu(s) Cu 2+ (aq) + 2 e - E = v 2 x red Ag + (aq) + e - Ag(s) E = v cell Cu(s) + 2 Ag + (aq) Cu 2+ (aq) + 2 Ag(s) E cell = v b) E cell [5 points] E cell = 0.46 v (see above) c) E cell [8 points] From Nernst E cell = E cell (RT/nF) ln(q) Q = [Cu 2+ ] = (0.5000) = 1.25 x 10 5 [Ag + ] 2 ( ) 2 E cell = (0.46 v) + (8.314 J/mol. K)(298. K) ln(1.25 x 10 5 ) = 0.61 v (2)(96485 C/mol) 10