Name:_____________
Chemistry 332
Final Exam
Calculations - Choose 5
1. In my research I have a compound that dissolves the plastic weighing boats. As a
result, I have to put a 100 ml volumetric flask on the analytical scale and add a few :l of
the material at a time until I get the right weight. Calculate the following molarity,
calculate the uncertainty in this molarity, and report the molarity with the correct
number of significant figures to reflect this uncertainty.
Weight of 100 ml volumetric
156.9837g
Weight of 100 ml volumetric + compound
157.0111g
Volumetric is a Class A (±.08 ml)
Molecular weight of compound is 159.629g/mol
Error in weight
157.0111+/- .0001
-156.9837 +/- .0001
= .0274 +/- sqrt(.00012 + .00012)
= .0274 +/- .00014 (absolute)
Calculation of Molarity
(g/molar mass)volume
(.0274/159.626)/.1 liter
= 1.71658x10-3 M
Multiplication/division calculation so need to use relative errors
error = sqrt[(.00014/.0274)2 + (.08/100)2]
.0052 (relative)
1.71658x10-3 x .0052 = 8.9x10-6 absolute
so the exact answer should be 1.716x10-3 +/- .009x10-3 M
Since sig fig assume +/- 1 in the last digit, and.009 is about = .01
I would prefer 1.72+/-.01 mM
2. In finding the molarity of Ca2+ in a water sample using EDTA I obtained the following
results over 6 runs.
42.18667 42.49333 46.64333
42.31 41.49333 42.79667
A. Can you justify eliminating the third value?
Q = gap/range
= (46.6433-42.79667)/(46.6433-41.4933)
=3.847/5.15 = .747
Qtable = .56 for 6 observations
Q>Qtable, value may be discarded
B. What is the mean and standard deviation of this distribution? (If you eliminate the
bad point do not include it in part B)
mean = 42.256 std dev = .48
2
C. When I used the Atomic Spectroscopy, in 5 runs I had an average value of 41.3 with
a standard deviation of .2. Statistically speaking, are these results significantly
different at the 95% probability level?
To do a t comparison you need to start with a pooled S
Spooled = sqrt[(.482(5-1) + .22(5-1))/(5+5-2)]
=.368
t=(42.256-41.3)/.368 × sqrt(5@5/(5+5))
= 4.1
t table for 10-2 or 8 degrees of freedom at the 95% level is 2.306
tcalc > ttable so numbers are significantly different
3. I am going to mix 10 ml of .1M AgNO3 with 10 ml of .1M Na2CrO4. Silver chromate
has a Ksp of 1.2x10-12. What are the concentrations of Na+, CrO42-, NO3- and Ag+ in the
resulting solution? What is the ionic strength of this solution?
In the first solution,10 ml x .1M = 1 mmole of Ag+ and 1 mmole of NO3In the second solution, 10 ml x .1M = 2 mmole Na+ and 1 mmole CrO42Ag+ and CrO4-2 will react to form a precipitate:
2 Ag+(aq) + 1 CrO42-(aq) 6Ag2CrO4(s)
Making a reaction table:
2 Ag+(aq) + 1 CrO42-(aq) 6Ag2CrO4(s)
Initial
1 mmole
1 mmole
0
Reaction
-1
-.5
+1
Net
0
.5
1
So before we do any equilibrium calculation you would expect all the silver to
precipitate out and half the chromate to remain in solution.
The molar concentrations before equilbrium are:
[Na+] = 2 mmoles/20 ml = .1M
[NO3-] = 1 mmole/20 ml = .05M
[Ag+] ~0
[CrO4-2] ~ .5 mmol/20 ml = .025M
There will be some silver due to the Ksp equilibrium
Ksp = 1.2x10-12 = [Ag+]2[CrO42-]
= X2(.025-X)
Let’s assume .025-X~.025
1.2x10-12 = X2(.025)
X = sqrt(1.2x10-12/.025)
= 6.92x10-6
.0000069 is small compared to .025 so our final numbers are:
[Na+] = 2 mmoles/20 ml = .1M
[NO3-] = 1 mmole/20 ml = .05M
[Ag+] = 6.92x10-6 M
[CrO4-2] ~ .5 mmol/20 ml = .025M
3
Ionic Strength
=(.1(+1)2 + .05(-1)2 + .7x10-6(+1)2 + .025(-2)2 )/2
=.125
4. I am going to mix 25 mls of .1M HCl with 10ml of .05M NaOH, what is the pH of the
solution?
1st solution 2.5 mmoles of H+
2nd solution .5 mmole of OHNet reaction will leave 2 mmoles of H+ remaining
2 mmole/35 mls = .0571 M H+
pH = -log(.0571) = 1.24
5. I am going to mix 25 mls of .1M Acetic acid with 10 ml of .05M NaOH, what is the pH
of the solution? (The Ka of acetic acid is 1.75x10-5)
2.5 mmole HOAc
.5 mmoles OHReaction Table:
HOAc + OH 6OAc- + H2O
Initial
2.5
.5
0
Reaction
-.5
-.5
+.5
Net
2
0
.5
Mixture of acid and conjugate base so use Henderson Hasselbalch
pH = 4.76 + log (.5/2)
= 4.16
6. I am going to mix 25 mls of .1M Na3PO4 with 10 ml of .05M H3PO4. What is the pH of
the resulting solution (H3PO4 has 3 pKa’s, 2.15,7.20, and 12.35)
25 mls x .1M = 2.5 mmoles Na3PO4 = PO4310 mls x .05 M H3PO4 = .5 mmoles H3PO4
We will assume the H3PO4 acts as a strong acid to react with the Na3PO4 acting
like a weak base.
Because you react one form of the phosphate to get another form, getting the
right reaction is a bit interesting. I started with the following reaction haveing the
triprotic acid react with the base:
H3PO4 + 3PO43-6 3HPO42- +PO43At the end of this reaction, however, the acid has created more base, so we can
get rid of this spectator and rebalance the equation
H3PO4 + 2PO43-6 3HPO42With this reaction, set up a reaction table:
H3PO4 +
2PO43-6
3HPO42Initial
.5
2.5
0
rxn
-.5
-1.0
+1.5
net
0
1.5
+1.5
4
So we have a mix of equal amounts of HPO42- and PO43-. Recognizing HPO42- as the
acid and PO43- as the base form in the last ionization of phosphoric acid you can solve
the pH using the Henderson-Hasselbach equation or you just use the pKa
pH = 12.35 + log (1.5/1.5)
= 12.35 + log (1)
= 12.35
7. I am going to mix 25 ml of .1M Ce4+ with 10 ml of 0.09M Tl1+
(Ce4+ + 1e- 6 Ce3+ ,Eo= 1.44; Tl3+ + 2 e- 6Tl1+, Eo = 1.22)
Calculate the potential of this solution vs. a saturated calomel electrode
I will start by ignoring the saturated calomel part, and just concentrate on the
standard potential vs an SHE. First figure out the reaction
The Ce has the highest potential so it will go forward, so your net reaction,
including electron stoichiometry must be: 2Ce4+ + Tl1+62Ce3+ + Tl3+
Finding the moles of starting reactants
25 ml x /.1M = 2.5 mmoles Ce4+
10 ml x .09M = .9 mmoles Tl1+
Setting up the reaction table:
2Ce4+ +
Tl1+6 2Ce3+ +
Initial 2.5
.9
0
rxn
-1.8
-.9
+1.8
Net .7
0
1.8
Tl3+
0
+.9
.9
You have non-zero amounts of Ce4+ and Ce3+ so this is the couple you use for to
calculate the potential. Setting up the Ce Nernst equation based on the half reaction:
E = 1.44 -.059/1 log ([Ce3+]/[Ce4+]
= 1.44 -.059 log [(1.8 mmole/35 ml)/(.7 mmole/35ml)]
= 1.4158 vs SHE
Now worry about the saturated calomel electrode
The SCE has a potential of +.241 vs SHE
so our potential vs SCE = 1.416-.241 = 1.175V
5
8. Sn has an emission line at 284 nm
A. What is the energy of this transition
E = h< = hc/8
= 6.626x10-34 x 3x108 / 284x10-9
= 6.999x10-19 J/photon
B. Say we have a plasma that is at 6000K, what percent of the Sn atoms in the
flame are at the high energy level so they can emit this light?
Assume g* is 3 and go is 2, the Boltzmann distribution equation
is:
Non-Calculations - Choose 4
No key provided
9. Diagram a galvanic cell that uses the reactions Hg2Cl2(s) + 2e- WHg(l) + 2Cl-, Eo =
.268) and AgCl(s) + e- WAg(s) + Cl-, Eo = .222) What is the potential of this cell if [Cl-] is
.01M in both cells.
10. I have a chemical reaction that must be performed at pH 4. I could make a solution
that is pH 4 by using .0004M HCl, or I could futz around with a mixture of acetic acid
and sodium acetate and make a buffer at pH 4. Why do chemists usually bother to
make a buffer?
11. Compare and contrast Flames, furnaces and Inductively Coupled Plasma as
methods to prepare atoms for flame emission and flame absorption experiments.
12. Calcium is not very soluble in the presence of CO32- (Ksp = 6x10-9), SO42- (Ksp = 2.4
x10-5) or hydroxide (Ksp = 6.5x10-6). I want to mix 5 mls each of .01M CaCl2, .01M
H2SO4, .01M NaOH, and .01M Na2CO3 together. Come up with as many equations as
possible that you could use to try to solve this system. DO NOT TRY TO SOLVE, just
get the equations. (Oh, yes you might also need the Ka’s of sulfuric acid, 4 and
1.03x10-2)
13. One thing we discovered, but I forgot to ask in a test, is to convert the Eo of a
reaction to either Keq and )G . In this equation you found that )G = -nFEo , so Eo must
be positive for a spontaneous reaction to occur.
Here is a related item that I know several you had problems with. When you do
potential calculations of redox titrations, the potentials you calculate can be + or
negative, depending on which reactant you are looking at and the Nernst equation of
that particular ½ reaction. Explain how you can have a spontaneous reaction and a
negative potential in a redox titration.