1 Page of 8
Spring 2007
Chem 208- Prelim 2
Name: ___________________________ Last First M.I. Lab: ___________________________ Day Time CU ID: ______________________
Lab TA Name: _________________
Instructions: Answer all questions in the spaces provided. You have 1 hours. You must show all work for full credit. The last page of this exam consists of an equation sheet, which may be detached if desired. There are a total of 8 pages in this exam. Graphing calculators and calculators that allow alphanumeric input are not allowed. There are 4 sections; the points for each is indicated below. 1.______________/20 2.______________/30 3.______________/25 4.______________/25
Total _____________/100
Page 2 of 8 1. Ammonium cyanide, NH4CN, dissociates completely in aqueous solution to form ammonium ions, NH4+(aq) and cyanide ions, CN- (aq). The following information (25oC) is provided: (20 pts) HCN: NH3: Ka = 4.9 x 10-10 Kb = 1.8 x 10-5
The most important reactions in ammonium cyanide solutions are: NH4+(aq) + H2O (l) CN- (aq)+ H2O (l) NH4+(aq) + CN-(aq) a) Calculate a numerical value for Kx. NH3(aq) + H3O+ (aq) HCN(aq) + OH- (aq) NH3(aq) + HCN (aq) Kx = ?
Ky = ?
Kz = ?
b) Calculate a numerical value for Ky.
c) Calculate a numerical value for Kz.
d) Will a 1.0 M aqueous NH4CN solution be basic, acidic, or neutral? Briefly explain your reasoning.
Page 3 of 8 2. a) What is the pH and percent ionization for a 1.500 M solution of methylamine, CH3NH2 (Kb = 4.3 x 10-4) at 25oC? (30 pts)
b) What is the pH of 0.750 M HCl at 25oC?
c) A student titrates 100.0 mL of the 1.500 M CH3NH2 solution with 0.750 M HCl. After 20.0 mL of the HCl has been added to the CH3NH2 solution, what is the pH?
Page 4 of 8 d) Continuing with the titration, what is the pH at the equivalence point?
e)
What is the pH of the solution after a total of 300.0 mL of the HCl solution has been added?
f)
The following indicators are available. Which would be the best choice for the above titration? Why? Indicator Bromphenol blue Chlorphenol Red Bromothymol Blue Thymolphthalein KHIn 1.4 x 10-4 1.0 x 10-6 7.9 x 10-8 1.0 x 10-10
Page 5 of 8 3. The Ksp of Fe(OH)3 is 4.0 x 10-38 at 25oC. (25 pts) a) What is the solubility of solid Fe(OH)3 in pure water, in moles/L?
b) What is the molar solubility solid of Fe(OH)3 in 1.00 x 10-3 M NaOH(aq)?
c) Will a precipitate of Fe(OH)3 form from a solution that is 0.013M Fe3+ in a buffer solution that is 0.150 M chlorous acid (HClO2) and 0.250 M sodium chlorate (NaClO2)? Note that Ka = 1.1 x 10-2 for HClO2.
[Fe(CN)6]3- (aq) is 1.0 x 1042. What is d) The formation constant, Kf, for Fe3+(aq) + 6CN (aq) the molar solubility of Fe(OH)3 in a 1.0 M aqueous KCN solution?
Page 6 of 8 4. (25 pts) The steam reforming of methane is an important industrial process for converting CH4 to CO: CH4(g) + H2O(g) CO(g) + 3H2(g) Kc = 0.260 at 1200K
a) A student mixes 1.00 mole of CH4 with 1.00 mole of H2O in a 1.0L flask at 1200K, and allows the reaction to come to equilibrium. What are the equilibrium concentrations of each component, in moles/L? Note that there is no need for successive approximations to do this problem.
b) What is the percent conversion of CH4(g) to CO(g) in part a?
Page 7 of 8 c) Starting with the equilibrium mixture in part a), if the following changes are made, will the equilibrium shift to the left, to the right, or remain unchanged? In each case, explain your answer in no more than one sentence: CO(g) + 3H2(g) Kc = 0.260 at 1200K CH4(g) + H2O(g) i. The total volume, V is increased.
ii.
The temperature is increased. For this reaction, Ho = 206.1 kJ.
iii.
Some platinum catalyst is added.
iv.
Some argon is added, causing the total pressure to double.
d) You evacuate a 1.0 L flask, and then add 1.00 mole of CO(g). How many moles of H2 (g) would you need to add to the flask so that at equilibrium, exactly 0.70 moles of CO(g) has reacted?
Page 8 of 8
Equation Sheet (may be removed if desired)
[A - ] pH = pK a + log10 [HA]
pH = -log10[H3O+] pOH = -log10[OH-] pH + pOH = 14.00 at 25oC Kw = [H3O+][OH-] = 1.00 x 10-14 at 25oC pKa = -log10Ka pKa + pKb = 14.00 for conjugate acid/base pairs at 25oC
log
x = log x - log y y
R = 8.31451 J mol-1K-1 R = 0.082057 L atm mol-1K-1 1 cal = 4.184 J NA = 6.02214 x 1023 mol-1 T(K) = T(oC) + 273.15 1 atm = 760 Torr = 101.3 kPa
Gas Constant:
Energy Conversion: Avogadro's Number: Kelvin - Celcius: Pressure: Equilibrium Const:
K p = K c (RT)
n gas
Quadratic Formula: Atomic Masses:
For ax + bx + c = 0, C = 12.01 O = 16.00 H = 1.01
2
- b b 2 - 4ac x= 2a