Name:
April 11, 2011
Chemical Principles II
Second Hourly Exam
1) When the question says to calculate or estimate, all calculations must be shown CLEARLY to
receive credit! 2)Answers must be filled in at end of each question
1
( 10 pts) Calculate the pH of a solution prepared by adding 0.080 mol nicotinic acid (NtH)
and 0.040 mol of sodium hydroxide to enough water to make a 2.00 L solution. pKa (NtH)=4.85
NtH + OH-  Nt- + H2O
Three component problem making a weak acid buffer.
0.080 mol NtH
0
mol Nt
0.040 mol OHmol HA - mol OH
[H+] = Ka -----------------------mol A + mol OH
mol HA - mol OH
0.080 mol NtH – 0.040 mol OH
pH = pKa – log ---------------------- = 4.85 – log -------------------------------------mol A + mol OH
0 mol Nt- + 0.040 mol OH
0.040 mol NtH
pH = 4.85 – log -----------------0.040 mol Nt-
= 4.85 – log 1 = 4.85
ANSWER = 4.85
2
(10 pts) Calculate the pH after 100.0 mL of 0.10 M HCl are added to a solution prepared
by adding 0.080 mol nicotinic acid (NtH) and 0.040 mol of sodium nicotinate (NaNt) to enough water
to make a 2.00 L solution.
1) Nt- + HCl  NtH + Cl2) calculate moles of the three components:
0.080 mol NtH
0.040 mol Nt
0.100 L x 0.10 MHCl = 0.010 mol HCl
3)
0.080 mol NtH + 0.010 mol HCl
0.090 mol NtH
pH = 4.85-log ---------------------------------------- = 4.85 – log ------------------ = 4.85 - log 3 = 4.37
0.040 mol Nt- - 0.010 mol HCl
0.030 mol Nt
ANSWER = 4.37
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3
(5 pts) (a)What is the concentration of punicate [PecO2-2] in 0.10 M punic acid H2PecO2?
Ka1 = 1.0 x 10-6, Ka2 = 1.0 x 10-11? (b)Calculate its pH.
(a) ANSWER
[PecO2-2] = Ka2 = 1.0 x 10-11
(b) ANSWER [H+] = sqrt (Ka1 x [H2PecO2] = sqrt (1.0 x 10-6,x 0.10 M) = 3.5
4 (3 pt) The pH of 100.0 mL of a buffer solution that is 0.25 M in HOAc and 0.25 M in OAc- is 4.74.
Ka (HOAc) = 1.8 x 10-5. What is the pH when 20.0 mL of water are added to this solution?
28ANSWER
= 4.74
5 (8 pts) The Kb = 1.8 x 10-5 for NH3(aq) NH4+(aq) + OH-(aq). Calculate the pH of 10.0 ml of a
0.10 M NH4Cl solution.
One component, Salt of weak base is acidic, Calculate [H+] from Ka
NH4+  NH3 + H+
Ka = Kw / Kb = 10-14 / 1.8 x 10-5 = 5.56 x 10-10
[H+] = sqrt( Ka . [NH4+]) = 5.56 x 10-10 x 0.10M NH4+ = 3.09 x 10-5
pH = - log(3.09 x 10-5) = 4.5
ANSWER = 4.5
6 (8 pts) Calculate the pH of 100 ml of a 0.010M sodium plastate solution, Ka (plastic acid,
HC3H5O3) = 1.0 x 10-4
One component, Salt of weak acid is basic, Calculate [OH-] from Kb
Kb = Kw / Ka = 10-14 / 1.0 x 10-4 = 1.0 x 10-10
[OH-] = sqrt(1.0 x 10-10 x 0.01M plastate) = 10-6
pOH = 6
pH = 14 – pOH = 8
44ANSWER
= 8.0
2
7 (2 pts) What is the criterion for “dropping x” in an equilibrium expression?
The equilibrium constant must be at least three orders of magnitude less than the concentrations.
8 (4 pt) Draw a circle around the stronger acid in each pair of compounds.
CF3NH2 and CF3OH,
H2O and H2S,
–
–
HSO4 and HPO4 ,
HClO3 and HClO2
10 (5 pt) Give the formulas and names of the strong acids.
HCl hydrochloric acid
HBr hydrobromic acid
HI hydroiodic acid
HNO3 nitric acid
H2SO4 sulfuric acid
HClO4 perchloric acid
HClO3 chloric acid
9 (2 pts) Identify any Lewis acid (LA) and Lewis base (LB) in the following reactions below. If
there is no acid-base reaction, put NAB after equation.
+3
+3
-1
3 Cl + Fe
+ 6 CO  [Fe(CO)6]
+ 3 (Cl )
Zn + 2 HCl  ZnCl2
+
H2
(LA):
Fe
+3
(LA): NAB (Ox-Red reaction)
(LB): CO
(LB):
57
3
Useful information: R = 8.314 J/mol or 0.08206 L•atm/mol•K
Substance Ho298-f kJ/mol So298-f J/mol.K
C4H10(g)
-126
201.5
H2(g)
Not given
130.6
C4H6 (g)
-26
192.5
10 ( 12 pts each) A process for the decomposition of butane involves the following equilibrium:
C4H10 (g)  C4H6 (g) + 2H2(g)
A
 B
2H
a)
Calculate HoSoGo and K for this reaction at 298K. Under these conditions, is
the reaction at equilibrium or will H2 be spontaneously formed? (You can use the abbreviations A, B,
H instead of writing the molecular formulas.)
Use ΔHo and So of formation from tables to calculate ΔHo and ΔSo of reaction.
ΔXo(reaction) = ( n x ΔXof PRODUCTS) - ( n x ΔXof REACTANTS)
ΔGo298 = ΔHo298 – 298K ΔSo298
K298 = e^(- ΔGo298 / RT)
Use R = 8.314 J/mol , T = 298K)
Answers: HoSoGo _____________ K _____________
b)
Estimate Go and K for this reaction at 500K. Under these conditions, is the reaction
at equilibrium, or will H2 spontaneously be formed when raising the temperature?
ΔGo500 ≈ ΔHo298 – 500K ΔSo298
K500 = e^(- ΔGo500 / RT)
Use R = 8.314 J/mol , T = 500K)
Answers: Go _____________ K _____________, formed?_____________
4
c)
Calculate G for this reaction at 500K and all gases with partial pressure = 10.0 atm.
Under these conditions, is the reaction at equilibrium or will more H2 tend to be formed when raising
all the partial pressures from 1.0 to 10.0 atm for this reaction?
ΔG = ΔGo + RT lnQ
Q = p(C4H6) . p2(H2) / p(C4H10) = 10 x 102 / 10 = 100
Answers: Go _____________ K__ same as b: K500___, formed?_____________________
11 ( 2 pts) What are the “standard conditions” that the symbol  refers to in the term G?
1 Molar contration and 1 barr pressure
The symbol  does not refer to the temperature, which is written as a subscript
The standard temperature is 298.15K.
12 (5 pts) (5 pts) Calculate the solubility product, Ksp of RaBr2 whose solubility is 1. x 10-6
mol/L.
“Solubility” means the number of moles of the entire substance (RaBr2) per L of solution.
x Mol RaBr2 = x mol Ra2+ = 2x mol BrKsp = [x][2x]2 = x . 4x2 = 4 x 3 = 4 . (1. x 10-6) = 4.0 . 10-18 M
Ksp_______________________________
Bonus section
BONUS
(5 pts) There are two methods to estimate the ΔGat another temperature than 298K. Use the other
method than you used in question 10b.
Van’t Hoff
Ln(KT1 / KT2) = - ΔHo/R . (1/T1 – 1/T2)
The First and Second Punic Wars were fought between Roman and Carthage.
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