Name:_____________
Chemistry 232
Third Hour Exam
Answer any 2 of the questions on this page. If you answer all three I will choose your best 2 for
your final score.
1. (10 points) I have a solution that is 1M Na3PO4. What is the ionic strength of this solution.
1M Na3PO4 will release 3 moles of Na+ and 1 mole of PO43- as it ionizes thus
[Na+] = 3M, [PO43-] = 1M, and if you want to get picky [H+]=[OH-]=1x10-7
:=1/2[3(1)2 + 1(-3)2 + .0000001(1)2 + .0000001(-1)2]
= (3+9)/2 (ignoring the H+ and OH- because their concentrations are so small
= 6M
2. (10 points) Use the equation log γ =
− .51z 2 µ
to determine the activity of Sn4+ in a solution with
α µ
305
an ionic strength of 0.1M. (The hydrated radius, ", of Sn+4 is 1100 pm)
1+
log γ =
− .51(4) 2 0.1
1 + 1100305 0.1
− 2.58
log γ =
= − 1.206
2.140
γ = 10 −1.206
γ = .0623
Activity = .0623 × [Sn + 4 ]
3. (10 points) Write both a charge balance and a mass balance equation for a .1M H2SO4. You can
assume the H2SO4 ionizes to both HSO4- and SO4-2.
Charge balance: [H+] = [HSO4-] + 2[SO42-] + [OH-]
Mass Balance 0.1M = [H2SO4] + [HSO42-] + [SO42-]
2
Answer any 2 of the questions on this page. If you answer all three I will choose your best 2 for
your final score.
4. (10 points) The formation constant (K f) for the Al-EDTA is 2x1016. Assuming the minimum K
conditional (Kf’) needed for a successful titration is 108, what is the lowest pH at which an EDTAAluminum titration can successfully take place?
Some values of "y-4 for EDTA
pH
"y-4
3
2.6x10-11
4
3.8x10-9
5
3.7x10-7
6
2.3x10-5
K’="y-4×Kf
K’$108
"y-4×Kf $108
"y-4×2x1016 $108
"y-4 $108/2x1016
"y- $5x10-9
pH 4 is below this so it does not work, pH 5 is above so it is OK
5. (10 points) I am titrating 25 mls of .00345M Cu+2 with .00456M EDTA at pH=7, where the Kf’ for
Cu+2 and EDTA is 3.15x1015. Calculate the pCu after I have added 25 mls of EDTA
1st where is the equivalence point?
1:1 Cu:EDTA complex so at the equivalence point MCuVCu=MEDTAVEDTA
25(.00345) = X(.00456
X=25(.00345)/.00456 = 18.91 ml
The chosen point is after the equivalence point so you use the equation
Kf’=[Cu+2 @EDTA]/[EDTA][Cu+2]
[Cu+2@EDTA] = moles of Cu+2/total volume
= 25(.00345)/50 = 1.725x10-3
[EDTA] = molarity EDTA ×(excess volume EDTA/total volume)
= .00456×(25-18.91)/50 = .5554x10-3
Technically the [Cu+2@EDTA] is 1.725x10-3 -X
And the [EDTA] = .5554x10-3 +X, but I will assume X is small and can be
ignored in these two concentrations hence:
3.15x1015 = 1.725x10-3/(.5554x10-3×[Cu+2])
[Cu+2]=1.725x10-3/(.5554x10-3×3.15x1015)
= 9.86x10-16
3
pCu=15.01
(Actually the third question from the previous page)
6.(10 points) Diagram a cell that uses the reactions Cd2+(aq) + 2e-YCd(s)
(Eo= -.402) and Hg2+ (aq)+ 2e-YHg(s) (Eo=.852) to generate a potential. Make sure the cell follows
the standard convention for electron flow, and on your diagram indicate: cathode, anode, ½ cell where
reduction occurs, ½ cell where oxidation occurs, where each chemical is in the cell, where the salt
bridge is.
Hard to diagram on a computer. Left cell has Cd+2 in solution and a Cd metal electrode, right
cell has Hg2+ in solution, a puddle of Hg metal on the bottom. And the electrode is inserted into the
liquid Hg. Electrons flow from left to right, the left electrode is called the anode and is where the
oxidation reaction occurs, the right electrode is the cathode and is where the reduction reaction occurs.
You should also be able to calculate the Eo of this cell, and make a line diagram of the cell, but I forgot
to ask that in this question
4
Answer any 2 of the questions on this page. If you answer all three I will choose your best 2 for
your final score.
7.(10 points) What is a formal potential, and why is it frequently used in Biology instead of a standard
potential
Formal potential refers to a potential derived under some non-standard state conditions. In Biology the
nonstandard state is pH=7, since that is where most of biology takes place
8.(10 points) Diagram and give the standard potential for and one standard reference electrodes
commonly used in a chemistry lab.
I was looking for a diagram of either a standard hydrogen electrode (Left half of figure 13-5 in
text, Eo=0.00) a silver/silver chloride electrode (figure 13-8, Eo= +.222 1M KCl or +.197 saturated
KCl) or a Calomel cell (figure 13-9, Eo=+.268 for 1M KCl, +.241 for saturated KCl)
9. (10 points) Diagram and explain how any one ion-selective electrode works.
Choose any ion selective electrode in chapter 14
5
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Take home portion
1. (20 points) In the lab you titrated 10 ml of .009M Ca2+ with .003M EDTA at pH 10. What was the
pCa in that titration at the following points: Initial, 15 ml of titrant, 30 ml of titrant and 45 ml of titrant
Where is the equivalence point?
EDTA-metal complexes are 1:1 so moles EDTA = moles Cu at equivalence point
MVEDTA=MVCa
.003(X)=10(.009)
X= 10(.009)/.003
X=30 ml
pCa = -log[Ca2+]
Initial
-log(.009) = 2.05
15 ml is before equivalence point, so [Ca+2] is simply the unreacted Ca in the total solution
[Ca2+] = [10(.009)-15(.003)]/(10+15)
= 1.8x10-3
pCa=2.74
30 ml is the equivalence point
In theory all the Ca2+ has been complexed with the EDTA, so the free Ca+2 comes from the
back reaction due to the equilibrium. Before we can tackle this problem you need to find the Kf’ for
this system
Kf’ = "Kf
= .36 × 1010.69
= 1.76x1010
1.76x1010 = [Ca@EDTA]/[Ca+2][EDTA]
[Ca@EDTA] = moles of Ca (or moles of EDTA)/volume
= 10(.009)/40
=2.25x10-3
[Ca2+] = [EDTA] = X
1.76x1010 = 2.25x10-3/X2
X2=2.25x10-3/1.76x1010
X=3.57x10-7; pCa=6.45
45ml
The chosen point is after the equivalence point so you use the equation
Kf’=[Ca+2 @EDTA]/[EDTA][Ca+2]
[Ca+2@EDTA] = moles of Ca+2/total volume
= 10(.009)/55 = 1.64x10-3
[EDTA] = molarity EDTA ×(excess volume EDTA/total volume)
= .003×(45-30)/55 = .818x10-3
7
Technically the [Ca+2@EDTA] is 1.64x10-3 -X
And the [EDTA] = .818x10-3 +X, but I will assume X is small and can be
ignored in these two concentrations hence:
1.76x1010 = 1.64x10-3/(.818x10-3×[Ca+2])
[Ca+2]=1.74x10-3/(.818x10-3×1.76x1010)
= 1.14x10-10
pCa=9.94
2. (10 points) I am going to make a cell. In the left hand ½ cell I am going to use a Ag electrode and
fill the cell with .05M Ag(NO3)2 . In the right had cell I am going to use a lead electrode and fill the cell
with .1M Pb(NO3)2. What is the potential of this cell? (It may or may not follow the standard
convention for electrons flowing from left to right)
Reaction right: Pb+2 + 2e-6Pb0
Er = -.126 -.059/2 log (1/.1)
Er = -.156V
Reaction left: Ag+ + 1e- 6Ag0
El = .799-.059/1 log (1/.05)
El = .722
E = Er-El = -.156-.722 =-.8775
3. (10 points) I am going to titrate Ag+ with Br-. I want to follow this titration with an experimental set
up similar to that shown on the margin of page 305. However, because I can’t find a saturated calomel
electrode in the lab, I will have to try to set up a standard hydrogen electrode for my reference cell.
What will the measured potential be if I start with 25 ml of .0050M Ag+, and I add 20 ml of .004M Br-.
This is a precipitation reaction. Is 20 ml of Br- before or after the equivalence point?
mMoles Ag+ = 25(.005) = .125
mMoles Br- = 20(.004) = .08
So I have an excess of Ag+
The Br- will make he Ag+ precipitate as AgBr, so the actual Ag+ remaining in solution
= (mmoles Ag+ - mmoles Br-)/volume
= (.125-.08)/(20+25)
= 1.00x10-3M
The system will respond directly to the Ag concentration using the Nernst equation
E=.799-.059/1 log (1/1x10-3)
E=.622V
and since the reference is the standard hydrogen electrode I don’t have to worry about any
additional offset in voltage