Name:____________
Chemistry 114
Fourth Hour Exam
1A(6 points) I want to make a solution that has a pH of 8.2. How many moles of NaOH will
I need to make 1 liter of this solution?
pH=8.2; pOH=14-8.2=5.8
5.8= -log[OH-]
-5.8=log[OH-]
10-5.8 = [OH-]
1.6x10-6 = [OH-] = [NaOH] so I need to add 1.6x10-6 moles of NaOH to 1 l of water
1B (7 points) I am going to make a solution that is .05M ammonia. What is the pH of this
solution? (The KB of ammonia is 1.8x10-5)
I give a Kb in this equation, so you don’t need to calculate it
At start
after rxn
NH3(aq) + H2O(l)6 NH4+(aq) + OH-(aq)
.05
Large
0
0
.05-X Large
X
X
Kb = 1.8x10-5 = [NH4+][OH-]/[NH3]
1.8x10-5 = X2/(.05-X) Assume .05-X..05
1.8x10-5 .X2/.05
X=sqrt(1.8x10-5×.05)
X = 9.5x10-5 = [OH-]=[NH4+]
pOH = -log(9.5x10-5)
=3.02
pH = 14-pOH = 10.98
2. (12 points) I am going to mix .5 moles of acetic acid (HC2H3O2) with 3.7 moles of sodium
acetate (NaC2H3O2) to make 2L of solution. What is the pH of this solution (The Ka of
acetic acid is 1.8x10-5).
pH = pKa + log(A-/HA)
pKa = -log Ka = -log(1.8x10-5) = 4.74
mole A- = 3.7
mole HA = .5
pH = 4.74 + log [(3.7/2)/.5/2)]
pH = 4.74 + log (3.7/.5)
pH = 4.74 +.87
pH = 5.61
2
3. (12 points) Class the following compounds as acids (A) bases (B) or neutral (N)
K2 O
__B__
KNO3 ___N_
NH4NO3 _A___
NO __A__
CaO __B__
Pb(NO3)4__A___
KHSO4 __A__
K2SO4 ___B_
4. (13 points) In chapter 22 you learned that aromatic compounds undergo substitution
reactions. An example from the text is shown below:
A proposed mechanism for this reaction is: Arrows sketched on test to show mechanism
don’t show up here
In the above mechanism:
Is compound A a Lewis acid or a Lewis base (circle one)
Is compound B a Lewis acid or a Lewis base (circle one)
Is compound C a Lewis acid or a Lewis base (circle one)
Is compound D a Lewis acid or a Lewis base (circle one)
Is compound E a Lewis acid or a Lewis base (circle one)
Is compound F a Lewis acid or a Lewis base (circle one)
3
5. Cu(IO3)2 has a solubility of 3.3x10-3 M.
A. (3 points) What is the concentration of Cu+2 in a saturated solution of Cu(IO3)2?
Cu(IO3)2 W Cu+2(aq) + 2IO3-(aq)
1:1 so for every
mole of-3 Cu(IO3)2 that dissolves, 1 mole of Cu+2 appears in solution
+2
[Cu ] = 3.3x10 M
B. (3 points) What is the concentration of IO3-in a saturated solution of Cu(IO3)2?
From the above equation, for every mole of Salt you get 2 moles of IO3[IO3-] = 2 × 3.3x10-3
= 6.6x10-3M
C. (3 points) What is the Ksp (solubility product) of Cu(IO3)2?
Ksp = [Cu+2][IO3-]2
=3.3x10-3 -7× (6.6x10-3)2
=1.77x10
D. (3 points) If I add Cu(NO3)2 to a saturated solution of Cu(IO3)2, will the solubility
of Cu(IO3)2 increase, decrease, or remain constant?
Common ion effect. Adding excess Cu+2 is like adding extra product, so
reaction will go to the left, more precipitate will form and solubility will decrease.
6. (13 points) The enthalpy of vaporization of mercury is 58.51 kJ/mol, and the entropy of
vaporization is 92.92 J/mol@K. What is the normal boiling point of mercury?
)Go = )H0 -T)So
At the normal boiling point liquid and vapor are in equilibrium.
If they are at equilibrium then )Go =0
0 = )H0 -T)So; )H0 =T)So
58510 = X(92.92) (converting )H to J so units are consistent)
X= 58510/92.92 = 630K or 357oC
4
7. Given the following data:
S(s) + 3/2 O2(g) - SO3(g)
2SO2(g) + O2(g) - 2SO3(g)
)G o= -371 kJ
)G o= -142 kJ
A (9 points) Calculate )G o for the reaction
S(s) + O2(g) - SO2(g)
A Hess’ law type problem
Can use 1st equation as is
Must reverse second equation and divide by 2
S(s) + 3/2 O2(g) - SO3(g)
)G o= -371 kJ
2/2 SO3(g) - 2/2 SO2(g) + ½ O2(g)
)G o= +142/2 kJ
)Go net = -371+ 71 = -300kJ
B (4 points) Assuming you are at 25o C use your answer for 7A, to calculate the Keq
for the same reaction. (If you don’t have an answer for 7A, just write down a number and
use it to get a corresponding Keq)
)Go = -RTlnK
= -8.314(J/K@mol) × 298K ×lnK
-300,000/(-8.314×298) = lnK
121=lnK
K = e121
K = 3.9x1052
8. (12 points) Define the following terms:
Positional Entropy - Entropy calculated based on the probability of finding a
particular state in a collection of different states
Free Energy )G, G = H -TS, The energy in a system available to do work.
A spontaneous process A process that will occur without outside intervention.
The second law of thermodynamics For any spontaneous process the entropy of
the universe always increases.
The third law of thermodynamics The entropy for a perfect crystal at 0K is 0.
A buffer A solution that resists changes to pH when either an acid or base is added
to it. Usually a mixture of an acid and its conjugate base, or a base and its conjugate acid.