Chemistry 321, fall 2011
Exam 0
1 September 2011
page 1
NAME: ______________________________ ID number: _____________________
This test has 7 questions and two extra credit questions. Answer all parts of the
questions. Write your final answer in the box at the bottom of each page. Express
all numeric answers to the correct number of significant figures.
1. A sample of pyrite (FeS) ore containing 36.40% S was decomposed and dissolved, then mixed
with a solution containing an excess of Ba(NO3)2. What weight of pyrite was used if the resulting
precipitate of BaSO4 weighed 1.0208 g? The molecular weight of BaSO4 is 233.42 g/mole. The
atomic weight of sulfur is 32.064 g/mole.
Answer: starting with the final product, 1.0208 g BaSO4 = (1.0208 g BaSO4)/(233.42 g/mole)
= 0.00437323279925 moles BaSO4
--> (1 mole S/1 mole BaSO4) --> 0. 00437323279925 moles S
g S = (00437323279925 moles S)(32.064 g/mole) = 0.140223336475 g S
Because this represents 36.40% of the weight of the sample, a simple ratio can be set up as
follows:
(wt of sample)/(100%) = (0.140223336475 g)/(36.40%)
wt of sample = (0.140223336475 g)(100)%/(36.40%)
= 0.38522894636 g = 0.3852 g
(4 sig figs because of 36.40%)
2. If 0.250 L of aqueous solution with a density of 1.00 g/mL contains 0.67 g of salt, express the
concentration of salt in wt %
Answer: mass of solution = (0.250 L)(1.00 g/ml)(1000 mL/L) = 250 g
Wt% = (mass of solute/mass of solution)(100) = (0.67/250)(100) = 0.268 % = 0.27% (2 sig figs)
3. Calculate the molarity of a 5.00 ppm Ca(NO3)2 solution. The molecular weight of Ca(NO3)2
is 164.09 g/mole
Answer: Since 5.00 ppm means 5.00 mg Ca(NO3)2/L solution, if mg can be converted to moles,
the problem is solved
molarity = (5.00 mg Ca(NO3)2 solution/L)(1 g Ca(NO3)2/1000 mg)
164.09 g/mole Ca(NO3)2
= 3.04710829423 x 10-5 M = 3.05 x 10-5 M (3 sig fig)
4. Mark the following compounds as soluble or insoluble.
a.
b.
c.
d.
K2CO3
Ca(ClO4)2
LiOH
Ag2S
soluble
soluble
soluble
soluble
insoluble
insoluble
insoluble
insoluble
Chemistry 321, fall 2011
Exam 0
1 September 2011
page 2
5. Given that a solution containing an unknown amount of sulfate (SO4=) is mixed with 50
mmoles of BaCl .2H 0 ( a soluble salt) resulting in the formation of solid BaSO .
2
2
4
a. If 35.7 mmoles of Ba2+ remains in solution and no SO4= remains dissolved, which species is
the limiting reactant in the formation of BaSO4.
Answer: If you started with 50 mmoles of Ba2+ and ended with 35.7 mmoles, you used up (50 35.7) = 14.3 mmoles by reacting it with an equivalent number of moles of sulfate. Because you
have excess Ba2+ in solution, the Ba2+ is not the limiting reagent-- the sulfate is the limiting
reagent. The ICE table looks like this:
+
SO42
BaSO4(s)
Ba2+
I
50.0 mmoles
?
0
C
-14.3 mmoles -14.3 mmoles
+14.3 mmoles
----------------------------------------------------------------------------------E
35.7 mmoles
0 mmoles
14.3 mmoles
This table shows that, since the reaction goes to completion (Keq = 1/Ksp), the amount of SO42(the limiting reagent) had to be 14.3 mmoles.
b. How much sulfate (in mmoles) was in the original solution.
50 – 35.7 mmoles = 14.30 mmoles = 14 mmoles (2 sig figs because of 50 mmoles)
6. Calculate the molar solubility of lead chloride (PbCl2) in pure water? The solubility product
constant (Ksp) for lead chloride is 1.0 x 10-5.
(a) Calculate the molar solubility of lead chloride (PbCl2) in a solution that is 0.050 M in calcium
Answer:
PbCl2(s) <===>
Pb2+ + 2 ClI
some
0
0
C
-s
+s
+2s
E
still some
s
2s
Ksp = 1.0 × 10-5 = s (2s)2 = 4 s3
s = (0.25 × 10-5)1/3 = 1.3572008808 x 10-2 M = 1.4 × 10-2 M
(b) Will the molar solubility of lead chloride (PbCl2) in a solution that is 0.050 M in calcium
chloride (CaCl2) be less or more than the solubility in pure water? Justify your answer with an
explanation. As above, the solubility product constant (Ksp) for lead chloride is 1.0 x 10-5.
Answer: The most straightforward answer and explanation is that, according to the ‘common ion
effect’ (as a consequence of Le Chatelier’s principle), the solubility of lead chloride (Ksp = 1.00
× 10-5) is greater in pure water than in a solution having an initial amount of the common ion
Cl-.
Chemistry 321, fall 2011
Exam 0
1 September 2011
page 3
The ICE calculation (not required) is:
0.050 M CaCl2 (a strong electrolyte)--100%--> 0.05M Ca+2, 0.10 ClI
C
E
PbCl2(s) <===>
some
-s
still some
Pb2+
0
+s
s
+
2 Cl0.10
+2s
0.10 + 2s
So, Keq = Ksp = 1.0 × 10-5 = s (0.10 + 2s)2
* assume 2s << 0.10
thus 1.0 × 10-5 = s (0.10)2
s = (1.0 × 10-5/(0.10)2) = 1.0 × 10-3
* check: % error = (100)(1.0 × 10-3)/0.10 = 1% so assumption is OK
Thus s = 1.0 × 10-3 M
7. Write net ionic reactions and the form of the resulting equilibrium constant expression (in
terms of Kw, Ksp, Ka, and Kb) for the following reactions:
(a) BaCl2 and Pb(NO3)2.
Pb2+(aq) + 2Cl- ・ PbCl2(s); Keq = 1/Ksp
(b) H2SO4 reacts with NaOH.
H+(aq) + OH-(aq) ・ H2O(l)l Keq = 1/Kw
(c) Ba(OH)2 + HCl
H+(aq) + OH-(aq) ・ H2O(l)l Keq = 1/Kw
(d) BaSO4 and HCl.
No reaction
Chemistry 321, fall 2011
Exam 0
1 September 2011
Bonus questions: (2 points each)
(a) Calculate the molarity of pure water? Show all your work.
H2O = 18 g/mole
D = 1 g /L
Assume 100g
Moles = 100 g / 18 g/mole = 5.556 moles
Volume = 100 g / 1 g/mL = 100 mL = 0.100 L
Molarity, M = 5.556 moels/0.100 L = 55.56 M
(b) What is the pH of a solution containing 1.2 x 10-8 M HCl?
HCl ---100%-->
1.2 x 10-8 M
H+_,
1.2 x 10-8 M
Cl1.2 x 10-8 M
pH = -log10(1.2 x 10-8 M) = 7.92
but you can’t produce a basic solution by acid to water, so…..
I
C
E
H20 <==>
-x
-
H+
+
-8
1.2 x 10 M
+x
1.2 x 10-8 M+ x
OH0
+x
x
Kw = [H+][OH-] = (1.2 x 10-8 + x)(x) = 1.0 10^-14
Solving this quadratic gives [H+] = 1.062 e-7, and pH = 6.97
page 4