Chm 152 Exam Review 2 (17, 18, 19, 20)

CHM152 Exam Review 2 (17, 18, 19 , 20) If an equal number of moles of HCN and KOH are added to water, what is the resulting solution: acidic, basic, or neutral? Explain. KOH is a strong base; HCN is a weak acid. When reacted together, they’ll produce water and KCN. K+ and CN- which will react with water (it is a weak base, a conjugate of a weak acid). This will produce a basic solution Determine the pH of a buffer solution that contains 4.0M CH3COOH and 3.0M NaCH3COO. Ka = 1.8x10-5. pH  pKa  log [conjugate base] [conjugate acid] pH  4.74  log 3.0M 4.0M = 4.62 Determine the pH of the buffer solution from above after you add 7.00mL of 1.0M HCl to 50.00ml of the buffer solution. 1.0mol HCl 0.00700L HCl  0.0070mol HCl L HCl HCl + NaCH3COO → CH3COOH + NaCl 4.0mol CH 3 COOH 0.05000L CH 3 COOH  0.20mol CH 3 COOH 0.0070mol 0.15mol 0.20mol L CH 3 COOH -0.0070mol -0.0070mol +0.0070mol 3.0mol NaCH 3 COO  0.15mol CH 3 COOH 0mol 0.143mol 0.207mol 0.05000L NaCH 3 COO L NaCH 3 COO 0.143mol pH  4.74  log = 4.58 0.207mol Determine the pH of the buffer solution from above after you add 4.00mL of 1.5M NaOH to 50.00ml of the buffer solution. 1.5mol NaOH 0.00400L NaOH  0.0060mol NaOH L NaOH NaOH + CH3COOH → NaCH3COO + H2O 0.0060mol 0.20mol 0.15mol 0.156mol pH  4.74  log -0.0060mol -0.0060mol +0.0060mol = 4.64 0.194mol 0mol 0.194 0.156 Calculate the pH of a 0.100 M NaCH3COO solution. Ka for acetic acid is 1.8 x 10-5. (Use A- as CH3COO-):6 i A- + H2O  HA + OH0.100 0 0 Kb = x2 / (0.100 – x) = 5.6 x 10-10 -x +x +x Assume x << 0.100 (and check it!) 0.100 – x - x Kb = x2 / 0.100 = 5.6 x 10-10 x2 = 5.6 x 10-11; x = 7.48 x 10-6; pOH = 5.13, pH = x 8.87 Calculate the pH of a 0.100 M CH3NH3Cl solution. Kb for methylamine, CH3NH2, is 3.7 x 10-4. (Use BH+ as CH3NH3+): BH+ + H2O  H3O+ + B 0.100 0 0 Ka = x2 / (0.100 – x) = 2.7 x 10-11 -x +x +x Assume x << 0.100 (and check it!) 0.100 – x x x Ka = x2 / 0.100 = 2.7 x 10-11 x2 = 2.7 x 10-12; x = 1.64 x 10-6; pH = 5.78 A 50.0 mL sample of 0.500 M HC2H3O2 acid is titrated with 0.150 M NaOH. Ka = 1.8x10-5 for HC2H3O2. Calculate the pH of the solution after the following volumes of NaOH have been added: a) 0 mL; b) 166.7 mL; c) 180.0 mL. a) 0 ml of base; only a weak acid is initially present so [H+] ≠ [HA] HC2H3O2 I C E - H+ + C2H3O2 0.500 -x 0.50-x 0 x x 0 x x  Ka = [ H  ][C 2 H 3 O2 ] [ HC 2 H 3 O2 ] 1.8x10-5 = x2 0.500 [H +] = x = 0.500( 1.8x10 5 ) = 3.0x10-3 pH = -log 3.0x10-3 = 2.52 b) 166.7 ml of NaOH are added  1 L  0.500 moles HC 2 H 3 O2     1000 mL  L   moles HC2H3O2 = 50.0 ml   1 L  0.150 moles NaOH    L   1000 mL  moles NaOH = 166.7 ml  - 2.50x10-2 moles HC2H3O2 2.50x10-2 moles NaOH - neutralization: HC2H3O2 + OH  C2H3O2 + H2O I C Final 0.0250 -0.0250 0 0.0250 -0.0250 0 only acetate remains – a weak base: - [C2H3O2 ] = 2.50 10 2 moles  0.115 M 0.2167 L 0 +0.0250 0.0250 - base hydrolysis: C2H3O2 + H2O I C E Kb for C2H3O2- = HC2H3O2 + OH 0.115 -x 0.115-x 0 x x 1  10 14 = 5.6x10-10 5 1.8 x10 Kb = - 0 x x [ HC2 H 3O2 ][OH  ] [C 2 H 3O2 ] 5.6x10-10 = x2 0.115 x = [OH-] =   0.115 5.6  10 10 = 8.0x10-6 pOH = -log 8.0x10-6 = 5.10 pH = 14 – 5.10 = 8.90  At the equivalence point for a WA/SB titration, the pH > 7 due to the OHproduced by the conjugate base hydrolysis reaction. c) 180.0 mL of NaOH are added from part b, moles HC2H3O2 = 2.50x10-2 moles HC2H3O2  1 L  0.150 moles NaOH  -2  moles NaOH = 180.00 ml   2.70x10 moles NaOH L   1000 mL  moles excess base = 2.70x10-2 moles - 2.50x10-2 moles = 2.0x10-3 moles NaOH M OH- = M NaOH = 2.0 x10 3 moles -3  8.7x10 M OH 0.2300 L pOH = -log 8.7x10-3 = 2.06 pH = 14 – 2.06 = 11.94 *Excess NaOH remains - this is the primary source of OH-. We can neglect the hydrolysis of the conjugate base because this would contribute a relatively small amount of OH- compared to the amount that comes directly from the excess NaOH. What is the concentration of a saturated silver acetate solution? Ksp(AgC2H3O2) = 1.94 x 10-3. Since Ksp = [Ag+][C2H3O2-], and the concentration of silver ions is the same as the concentration of acetate ions, we can set up the following equation: 1.94 x 10-3 = x2 x = 0.0440 M What is the concentration of a saturated lead chloride solution? Ksp(PbCl2) = 1.17 x 10-5. Ksp = [Pb+2][Cl-]2. Since the concentration of chloride ions is twice that of lead (II) ions, this boils down to the following equation: 1.17 x 10-5 = (x)(2x)2 1.17 x 10-5 = 4x3 Silver phosphate, Ag3PO4, is an insoluble salt that has a Ksp = 1.3 x 10-20. a) Calculate the molar solubility of Ag3PO4 in pure water. Ag3PO4(s) 3Ag+(aq) + PO43-(aq) Ksp = [Ag+]3[PO43-] I C E 0 3x 3x 0 x x Ksp = (3x)3x 1.3x10-20 = 27x4 x4 = 4.8x10-22 x = 4.7x10-6 M = molar solubility of Ag3PO4 in pure water b) Calculate the molar solubility of Ag3PO4 in a solution containing 0.020 M Na3PO4 (a soluble salt). soluble salt: Na3PO4  3Na+ + PO43Phosphate is the common ion: [PO43-] = [Na3PO4] = 0.020 M (since 1 mol Na3PO4 forms 1 mol PO43- ions) Ag3PO4(s) 3Ag+(aq) + PO43-(aq) I C E 0 3x 3x 0.020 x 0.020+x Ksp = [Ag+]3[PO43-] 1.3x10-20 = = (3x)30.020 6.5x10-19 = 27x3 x3 = 2.4x10-20 x = 2.9x10-7M = molar solubility of Ag3PO4 with a common ion  Adding common ion decreases the solubility of Ag3PO4 c) Will a precipitate form if a solution is made to contain 3.0x10-5M AgNO3 and 2.3x104 M Na3PO4? (Hint: Q vs. K) We already know Ksp and Qsp = [Ag+]3[PO43-] so we can plug the concentrations in to calculate Qsp. Qsp = [3.0x10-5]3[2.3x10-4] = 6.21x10-18. From above, Ksp = 1.3x10-20. Since Q>K, the reaction should shift to the reactant side (remember, the reaction was written with the solid as a reactant), and a solid will form. If Q had been smaller than K, a precipitate would not have formed. a. Calculate the standard free energy change, ΔG, for the following at 25 C: MgO(s) + C(graphite)  Mg(s) + CO(g) ΔH = 491.18 kJ G = H - TS ΔS = 197.67 J/K J   1 kJ    = + 432.27 kJ G = 491.18 kJ - (298 K) 197.67   K   1000 J   b. Is this reaction spontaneous at 25 C? If not, at what temperature can we make this reaction spontaneous? No, it is not spontaneous at 25 C (G is a large positive value) Set G = 0: T= 0 = H - TS  491.18 kJ = 2484.8 K J  1 kJ    197.67  K  1000 kJ   T= H S  This is T at equilibrium, spont at T > 2484.8 K For the unbalanced reaction 2 SO2(g) + O2(g) → 2 SO3(g) calculate ΔG at 25.0ºC when the reactants and product are at the following partial pressures: 10.0 atm SO2 , 10.0 atm O2 , and 1.00 atm SO3. G = Go + RT ln Q Q = 12 / (102 * 10) = 1/1000 Go = [ 2mol(-371.1 kJ/mol)] – [ 2mol(-300.2 kJ/mol) + 0 ] = -141.8 kJ G = -141.8 kJ + (8.314 J/mol·K) (298 K) ( kJ/1000J) (ln 1/1000) = -159 kJ In each of the following equations, indicate the element that has been oxidized and the one that has been reduced. You should also label the oxidation state of each before and after the process: 2 Na + FeCl2  2 NaCl + Fe  Sodium is oxidized, going from a 0 to +1 oxidation state.  Iron is reduced, going from a +2 to 0 oxidation state. 2 C2H2 + 5 O2  4 CO2 + 2 H2O  Carbon is oxidized, going from a –1 to +4 oxidation state.  Oxygen is reduced, going from a 0 to –2 oxidation state. 2 PbS + 3 O2  2 SO2 + 2 PbO  Sulfur is oxidized, going from a –2 to +4 oxidation state.  Oxygen is reduced, going from a 0 to –2 oxidation state. A galvanic cell is constructed based on the following reactions: Zn2+ + 2e-  Zn E red = -0.76 V Al3+ + 3e-  Al E red = -1.66 V a) Write the overall balanced equation for the cell and calculate the cell emf under standard conditions. 3(Zn2+ + 2e-  Zn) E red = -0.76 V 2(Al  Al3+ + 3e-) E ox = +1.66 V 3Zn2+ + 2Al  3Zn + 2Al3+ Eocell= -0.76 V + 1.66 V = 0.90 V b) Calculate G° at 298 K.  96500 J   0.90 V = -5.2 x105 J or -520 kJ G = -nFE = -(6 mol e-)  V  mol e    c) What is the value of the equilibrium constant, K, for this reaction at 298 K? Eo = (RT/nF) ln K; n = 6; K = 1.6 x 1091 Which combination below will undergo a spontaneous reaction? Half Reaction Table E red – Br2(l) + 2e-  2Br (aq) +1.06 V 2+ Cu (aq) + 2e-  Cu(s) +0.34 V Ni2+(aq) + 2e-  Ni(s) −0.28 V 3+ Al (aq) + 3e-  Al(s) −1.66 V a. b. c. d. e. Ni 2+(aq) with Br–(aq) Cu(s) with Al3+(aq) Ni (s) with Br–(aq) Br–(aq) with Cu2+(aq) Br2(l) with Ni(s) A voltaic cell is constructed that used the following reaction and operates at 298 K: 2Al(s) + 3Mn2+(aq)  2Al3+(aq) + 3Mn(s) a. What is the EMF of this cell under standard conditions? (Use Appendix D for Eored values.) +1.66 V + -1.18 V = 0.48 V b. What is the EMF of this cell when [Mn2+] = 0.10 M and [Al3+] = 1.5 M? E = 0.48 V – (0.0592 V / 6 mol e-) (log (1.5)2/(0.10)3) = 0.45 V 16. Give the symbol, atomic number, and mass number from each of the following:  beta particle :  alpha particle: 0 1  or since it is now know to be an electron 4 2  0 -1 e or since it is now know to be a helium nucleus 4 2 He   gamma radiation:  positron: the anti - particle to the electron,  neutron: 1 0 0 0 0 1 e n 17. Uranium-238, 238U, can undergo beta decay, where two beta particles are released. Write the equation for this nuclear reaction: 238 92  U → 2 10  + 238 94 Pu The same isotope can also undergo alpha decay, releasing one alpha particle. Write the equation for that mode of decay. 238 92 U → 4 2  + 234 90 Th 18. Another isotope of uranium, 229U, has a half-life of 58 minutes. Calculate the time, in hours, it would take for only 3.0% of a 100.0g sample to remain. Since nuclear decay follows first-order kinetics, we can use the same formulas: Instead of concentrations, we can also use mass or the number of disintegrations (a measure of radioactivity). Just depends on the information given. ln2 t 1/2  k 58 min  ln2 k k ln2  0.012 min -1 58 min  3.0%  -1 ln    (0.012 min ) t  100.%  Since it’s 1st order, we can just use % in the equation. For second order, we would use the actual masses. t  292min 292 min 1 hr  4.9 hr 60 min  [A ]  ln  t   kt  [A 0 ] 