Chapter 16: Acid-Base Equilibria and Solubility Equilibria

Transcription

1 STUDY-GUIDE: FOR TEST-3 CHEM 1412 Chapter 16: Acid-Base Equilibria and Solubility Equilibria A table of ionization constants and Ka 's is required to work some of the problems in this chapter [1]. Which of the following yields a buffer solution when equal volumes of the two solutions are mixed? A) M H3PO4 and 0.050M HCl B) 0.050M H3PO4 and M HCl C) 0.050M NaH2PO4 and 0.025M NaOH D) 0.050M Na3PO4 and 0.050M M NaOH E) 0.050M Na3PO4 and 0.025M NaOH Ans: C [2] Which of the following is the most acidic solution? A) 0.10 M CH3COOH and 0.10 M CH3COONa B) 0.10 M CH3COOH C) 0.10 M HNO2 D) 0.10 M HNO2 and 0.10 M NaNO2 E) 0.10 M CH3COONa Ans: C [3] You have ml of a buffer solution containing 0.30 M acetic acid (CH3COOH) and 0.20 M sodium acetate (CH3COONa). What will the ph of this solution be after the addition of 20.0 ml of 1.00 M NaOH solution? [Ka = ] A) 4.65 B) 4.71 C) 4.56 D) 4.84 E) 5.07 Ans: B

2 [4] ml of 0.10 M HNO2 (nitrous acid, Ka = ) is titrated with a 0.10 M KOH solution. After ml of the KOH solution is added, the ph in the titration flask will be Ans: B A) 2.17 B) 3.35 C) 2.41 D) 1.48 E) 7.00 [5] The molar solubility of manganese(ii) carbonate is M. What is Ksp for this compound? Ans: D A) B) C) D) E) [6] A saturated sodium carbonate solution at 100 C contains 45.5 g of dissolved sodium carbonate per 100. ml of solution. The solubility product constant for sodium carbonate at this temperature is Ans: E A) 79.0 B) C) D) 36.8 E) 316 [7] Solid sodium iodide is slowly added to a solution that is M Pb and M Ag. [Ksp (PbI2) = ; Ksp (AgI) = ] Calculate the Ag concentration when PbI2 just begins to precipitate. Ans: M [8] NaCl is added slowly to a solution that is M each in Cu+, Ag+, and Au+. The Ksp's for CuCl, AgCl, and AuCl are , , and , respectively. Which compound will precipitate first? Ans: AuCl 2

3 [9] The percent ionization of a weak acid HA is greater in a solution containing the salt NaA than it is in a solution of the weak acid only. Ans: False [10] What mass of ammonium nitrate must be added to 350. ml of a M solution of ammonia to give a buffer having a ph of 9.00? (Kb(NH3) = ) Ans: A A) 7.6 g B) 2.4 g C) 5.4 g D) 11 g E) 3.3 g [11]. In which one of the following solutions will acetic acid have the greatest percent ionization? Ans: A A) 0.1 M CH3COOH B) 0.1 M CH3COOH dissolved in 1.0 M HCl C) 0.1 M CH3COOH plus 0.1 M CH3COONa D) 0.1 M CH3COOH plus 0.2 M CH3COONa [12]. Which of the following yields a buffer solution when equal volumes of the two solutions are mixed? Ans: C A) 0.10M HCl and 0.10 M NaCl D) 0.10M HClO4 and 0.10 NaClO4 B) 0.15 M HNO3 and 0.15 M NaNO3 E) 0.15M HBr and 0.15M NaBr C) 0.10M HF and 0.10M NaF 3

4 [13]. Calculate the ph of a buffer solution that contains 0.25 M benzoic acid (C6H5CO2H) and 0.15M sodium benzoate (C6H5COONa). [Ka = for benzoic acid] Ans: A A) 3.97 B) 4.83 C) 4.19 D) 3.40 E) 4.41 [14]. You are asked to go into the lab and prepare an acetic acid - sodium acetate buffer solution with a ph of 4.00 ± What molar ratio of CH3COOH to CH3COONa should be used? Ans: D A) 0.18 B) 0.84 C) 1.19 D) 5.50 E) 0.10 ======================================================================= Chapter 17: Entropy, Free Energy, and Equilibrium ======================================================================= [15] Which of these species would you expect to have the lowest standard entropy (S )? Ans: E A) Br2(l) B) Cl2(g) C) F2(g) D) H2(g) E) I2(s) [16] Arrange the following substances in the order of increasing entropy at 25 C. HF(g), NaF(s), SiF4(g), SiH4(g), Al(s) lowest highest Ans: C A) SiF4(g) < SiH4(g) < NaF(s) < HF(g) < Al(s) B) HF(g) < Al(s) < NaF(s) < SiF4(g) < SiH4(g) C) Al(s) < NaF(s) < HF(g) < SiH4(g) < SiF4(g) D) Al(s) < HF(g) < NaF(s) < SiF4(g) < SiH4(g) E) NaF(s) < Al(s) < HF(g) < SiF4(g) < SiH4(g) 4

5 [17] Which one of the following reactions would you expect to have highest S? Ans: D A) CH4(g) + 2O2(g) CO2(g) + 2H2O(g) B) C2H2(g) + 5/2O2(g) 2CO2(g) + H2O(g) C) C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(g) D) C2H6(g) + 7/2O2(g) 2CO2(g) + 3H2O(g) [18] Sulfur can be separated from lead in the mineral galena, PbS(s), by roasting the ore in the presence of oxygen as shown in the following reaction: 2PbS(s) + 3O2(g) 2PbO(s) + 2SO2(g) Calculate S for this reaction using the thermodynamic data provided below Compound S (J/K mol) PbS(s) 91.2 O2(g) PbO(s) SO2(g) Ans: B A) 410 J/K mol D) 21.8 J/K mol B) J/K mol E) 43.5 J/K mol C) 47.7 J/K mol [19] With respect to the system only, a reaction with H > 0 and S < 0 is predicted to be: Ans: D A) Spontaneous at all temperatures B) Spontaneous at high temperatures only C) Spontaneous at low temperatures only D) Nonspontaneous at all temperatures 5

6 [20] At 1500 C the equilibrium constant for the reaction CO(g) + 2H2(g) CH3OH(g) has the value Kp = Calculate G for this reaction at 1500 C. Ans: E A) 105 kj/mol D) 105 kj/mol B) 1.07 kj/mol E) 233 kj/mol C) 233 kj/mol [21] Calculate Kp at 298 K for the reaction SO2(g) + NO2(g) SO3(g) + NO(g). G f SO2(g) kj/mol SO3(g) kj/mol NO(g) 86.7 kj/mol NO2(g) 51.8 kj/mol Ans: E A) B) C) 14.2 D) 475 E) [22] For the reaction SbCl5(g) SbCl3(g) + Cl2(g), G f (SbCl5) = kj/mol G f (SbCl3) = kj/mol H f (SbCl5) = kj/mol H f (SbCl3) = kj/mol Will this reaction proceed spontaneously at 298 K and 1 atm pressure? Ans: No 6

7 [23] Consider the reaction CO(g) + 2H2(g) CH3OH(l) at 25 C. G f (CO) = kj/mol G f (CH3OH) = kj/mol H f (CO) = kj/mol H f (CH3OH) = kj/mol S (CO) = J/K mol S (CH3OH) = J/K mol Calculate value of the equilibrium constant (Kp) for this reaction at 25 C. Ans: [24] Given the following data, calculate the boiling point of HCOOH (formic acid) H f (kj/mol) S (J/K mol) HCOOH(l) HCOOH(g) Ans: 115 C [25] Using the thermodynamic data provided below, calculate Ka for HCN(aq) at H f (kj/mol) S (J/K mol) H + (aq) 0 0 CN (aq) HCN(aq) Ans: (compared to the experimental value of ) 7

8 ========================================================================= Chapter 18: Electrochemistry NOTE: A table of standard reduction potentials is required to work many of these problems. ======================================================================== [26] What is the total number of moles (n) of electrons exchanged between the oxidizing agent and the reducing agent in the overall redox equation: 5 Ag + (aq) + Mn 2+ (aq) + 4H2O(l) 5Ag(s) + MnO4 (aq) + 8H + (aq) Ans: D A) 1 B) 2 C) 3 D) 5 E) 7 [27] Complete and balance the following redox equation. What is the coefficient of H2O when the equation is balanced with the set of smallest whole-number coefficients? Ans: A H2O + MnO 4 + I MnO2 + IO3 (basic solution) A) 1 B) 2 C) 4 D) 10 E) None of these. [28] Given the following cell diagram, Al(s) Al (aq) Cd (aq) Cd (s) what is the balanced overall (net) cell reaction? Ans: C A) Al(s) + Cd (aq) Al (aq) + Cd (s) B) Al (aq) + Cd (s) Al(s) + Cd (aq) C) 2Al(s) + 3Cd (aq) 2Al (aq) + 3Cd (s) D) 2Al (aq) + 3Cd (s) 2Al(s) + 3Cd (aq) E) 3Al (aq) + 2Cd (s) 3Al(s) + 2Cd (aq) 8

9 [29] A galvanic cell has the overall reaction: 2Fe(NO3)2(aq) + Pb(NO3)2(aq) 2Fe(NO3)3(aq) + Pb(s) Which is the half reaction occurring at the cathode? Ans: D A) NO (aq) + 4H (aq) + 3e- NO(g) + 2H2O(l) B) NO (aq) + 2H (aq) + e- NO2(g) + H2O(l) C) Fe (aq) Fe (aq) + e- D) Pb (aq) + 2e- Pb(s) E) Fe (aq) + e- Fe (aq) [30] Calculate E cell for a silver-aluminum cell in which the cell reaction is Al(s) + 3Ag (aq) Al (aq) + 3Ag(s) A) 2.46 V B) 0.86 V C) 0.86 V D) 2.46 V E) none of these Ans: D [31] Consider the following electrochemical cell: U U (aq) Cl (aq),cl2(g) Pt If the standard cell emf is 3.16 V, what is the standard reduction potential for uranium? A) 3.16 V B) V C) 1.80 V D) V E) V Ans: C 9

10 [32] Consider the following standard reduction potentials in acid solution: E (V) Al + 3e Al(s) 1.66 AgBr(s) + e Ag(s) + Br Sn + 2e Sn Fe + e Fe The strongest oxidizing agent among those shown above is Ans: A A) Fe. B) Fe. C) Br. D) Al. E) Al. [33] For the electrochemical cell Pt(s) H2(1 atm) H (1 M) Cu (1 M) Cu(s), which one of the following changes will cause an increase in the cell voltage? Ans: C A) Lower the H2(g) pressure. B) Increase the size/mass of the copper electrode. C) Lower the H (aq) concentration. D) Decrease the concentration of Cu ion. E) None of the above. 10

11 [34] Consider an electrochemical cell involving the overall reaction 2AgBr(s) + Pb(s) Pb + 2Ag(s) + 2Br Each half-reaction is carried out in a separate compartment. The anion included in the lead half-cell is NO3. The cation in the silver half-cell is K. The two half-cells are connected by a KNO3 salt bridge. If [Pb ] = 1.0 M, what concentration of Br ion will produce a cell emf of 0.25 V at 298 K? Given: AgBr(s) + e Ag + Br, E = V. A) 0.02 M B) 0.14 M C) 0.38 M D) 1.0 M E) 7.0 M Ans: B [35] How many coulombs (C) of electrical charge must pass through an electrolytic cell to reduce 0.44 mol Ca ion to calcium metal? Ans: B A) 190,000 C B) 85,000 C C) 21,000 C D) 42,500 C E) 0.88 C [36] How long will it take to produce 78 g of Al metal by the reduction of Al3+ in an electrolytic cell with a current of 2.0 A? Ans: D A) 0.01 s B) 420 s C) 13 h D) 116 h E) s [37] Which element is associated with the term "galvanized"? A) Ga B) Zn C) Cd D) Hg E) Pb Ans: B 11

12 [38] What is the total number of moles (n) of electrons exchanged between the oxidizing agent and the reducing agent in the overall redox equation: Ans: 6 3C2O4 (aq) + Cr2O7 (aq) + 14H (aq) 6CO2(g) + 2Cr (aq) + 7H2O(l) [39] How many grams of copper are deposited on the cathode of an electrolytic cell if an electric current of 2.00 A is passed through a solution of CuSO4 for a period of 19.0 min? Ans: g [40] Many different ways have been proposed to make batteries. One cell is set up with copper and lead electrodes in contact with CuSO4(aq) and Pb(NO3)2 (aq), respectively. If the Pb and Cu concentrations are each 1.0 M, what is the overall cell potential? Pb + 2e Pb Cu + 2e Cu E = 0.22 V E = V Ans: 0.56 V [41] The electrochemical cell that utilizes the reaction Zn + Cu (1 M) Zn (1 M) + Cu will have a lower cell emf when the concentrations of Zn and Cu ions are decreased to 0.1 M. Ans: False 12

13 [42] A metal object is to be gold-plated by an electrolytic procedure using aqueous AuCl3 electrolyte. Calculate the number of moles of gold deposited in 3.0 min by a constant current of 10. A. Ans: A A) mol D) mol B) mol E) 160 mol C) mol 13

14 CHEM 1412 Formulas and Constants mass of proton = amu mass of neutron = amu c = 3.00 X 10 8 m/s F = C/(mol of e - ) = J/(V mol of e - ) K = o C R = (L atm)/(mol K) = J/(mol K) 1 g = X amu 1 atm = 760 mm Hg S g = kp g P A = X A P A o T f = K f m T b = K b m = MRT PV = nrt [A] t = -kt + [A] o t 1/2 = [A] o / 2k ln ([A] t /[A] o ) = -kt ln[a] t = -kt + ln[a] o t 1/2 = / k 1/[A] t = kt + 1/[A] o t 1/2 = 1 / (k[a] o ) k = Ae -Ea/RT ln(k 1 /k 2 ) = E a /R (1/T 2-1/T 1 ) =================================================================== For the chemical reaction (at Equilibrium): aa + b B c C + d D Kc = [C] c [D] d [A] a [B] b (Law of mass action. General expression of equilibrium constant) Kp = Pc c PD d PA a PB b (Kp: when equilibrium concentrations are expressed in terms of partial pressures) Kp = Kc (0.0821*T) n, Relationship between Kp and Kc. n = moles of gaseous products moles of gaseous reactants. Kc = Kc Kc, The equilibrium constant for the overall reaction is given by the product of the equilibrium constants for the individual reactions. Reaction Quotient (Qc) when instead of Equilibrium concentrations, the initial concentrations are used in the equilibrium constant expression. Qc determines the direction in which the net reaction will proceed: (Qc < Kc: system proceeds from left to right, consuming reactants, forming products) (Qc = Kc: system is at equilibrium) (Qc > Kc: system proceeds from right to left, consuming products, forming reactants). =========================================================================== 14

15 =============================================================== K w = [H + ] [OH ] Ion-product constant of water. ph = - log [H + ], Definition of ph of a solution. [H + ] = 10 -ph Calculating H+ ion concentration from ph. poh = - log [OH ], Definition of poh of a solution. [OH ] = 10 -poh Calculating OH ion concentration from poh. ph + poh = 14 [H + ][OH - ] = 1.0 X K a K b = K w = 1.0 X Relationship between the acid and base ionization constants of a conjugate acid-base pair. Calculating ph of a solution containing a weak acid and a soluble salt of a weak acid (e.g. NaA): or HA(aq) + H2O (l) HA(aq) + H2O (l) H3O + (aq) + A - (aq) H + (aq) + A - (aq) The ionization constant Ka : K a = [H + ] [A - ] [HA] pk a = - log K a, Henderson-Hasselbach equation: ph = pk a + log [conjugate base] [conjugate acid] Percent ionization = ionized acid concentration at equilibrium 100% initial concentration of acid Ka = acid ionization (dissociation) constant (measure of a strength of an acid) Ka = [H3O + ] [A ] [HA]. 15

16 Kb = base ionization (dissociation) constant (measure of a strength of a base). Kb = [BH + ] [OH ] [B]. The net direction of an acid-base reaction proceeds to the greater extent in the direction in which a stronger acid and stronger base form a weaker acid and weaker base. Quadratic equation: a x 2 + b x + c = 0 x = b b 2 4 a c 2a ph = pk a + log([base]/[acid]) G o = H o - T S o G o = -RT ln K eq G = G o + RT lnq G o = -n F E o E = E o - (0.0592/n) log Q E = mc 2 Coulombs = Amps X seconds Soluble Compounds Important Exceptions Compounds containing NO 3 none Compounds containing C 2 H 3 O 2 none Compounds containing Cl Salts of Ag +, Hg 2+ 2, Pb 2+ Compounds containing Br Salts of Ag +, Hg 2+ 2, Pb 2+ Compounds containing I Salts of Ag +, Hg 2+ 2, Pb 2+ 2 Compounds containing SO 4 Salts of Ca 2+, Sr 2+, Ba 2+, Hg 2+ 2, Pb 2+ Insoluble compounds Important Exceptions Compounds containing S 2 2 Compounds containing CO 3 3 Compounds containing PO 4 Compounds containing OH Salts of ammonium, alkali metal cations and Ca 2+, Sr 2+, Ba 2+ Salts of ammonium, alkali metal cations Salts of ammonium, alkali metal cations Salts of ammonium, alkali metal cations and Ca 2+, Sr 2+, Ba 2+ Percent by mass = mass of solute 100% [mass of solute + mass of solvent] 16

17 Two-component solution: Mole fraction of solute A: XA = moles of solute A moles of solute A + moles of solvent B Molarity = number of moles of solute in 1 L of solution: Molarity = moles of solute Liters of solution Molality = number of moles of solute dissolved in 1 kg (1000 g) of solvent: Molality = moles of solute Mass of solvent (1 kg) Density = Mass / Volume. 17