Name:_____________
Chemistry 232
Third Hour Exam
1. ( 10 points) What is the ionic strength of a solution that is 0.1M FeCl2 and
.05M Al(NO3)3?
Solution will be .1M Fe+2, .2M Cl-1, .05M Al+3, .15M NO3Ionic strength = (.1(2)2 + .2(-1)2 + .05(+3)3 + .15(-1)2)/2
= .6M
2. (10 points) The extend Debye Huckel equation relating activity
coefficients to ionic strength is:
Assume the ionic strength of a solution is .05M
Calculate the activity of the ions NH4+ ("=250) and PO4-3("=400). Assuming
the molar concentration of these ions is .03 and .01 respectively.
NH4+
Activity NH4+ = .801×.03 = .0240M
PO4-3
Activity PO4-3 = .622×.01 = .0622M
2
3. (10 points) The log Kf for the Fe+2 EDTA complex is 14.3. If I mix 25 mls
of .05M EDTA and 35 mls of .025M Fe+2, what is the concentration of free
Fe+2 in the solution?
25 ml of .05M EDTA = 1.25 mmoles EDTA
35 ml of .025M Fe+2 = .875 mmoles Fe+2
Excess EDTA so past the equivalence point
Moles Fe@EDTA = .875
Moles Free EDTA = 1.25-.875 = .375mmoles
Log Kf = 14.3, Kf = 1014.3 = 2.00x1014
Kf = [Fe@EDTA]/[Fe][EDTA]
[Fe]=[Fe@EDTA]/Kf [EDTA]
= .875/[(2x1014).375] (Volumes cancel out)
=1.17x10-14
3
4. (10 points) I want to make a battery out of exotic materials because I think
I can sell it if it has a strange sounding name. Thus I want to use the
following two half reaction to make my Indium/Yttrium battery:
Yttrium
Y+3(aq) + 3e- W Y(s)
Eo = -2.38
Indium
In+3(aq) + 2e- W In+(aq)
Eo = -.444
Following standard conventions make a diagram of this cell. Indicate in this
diagram the + pole, the - pole, the anode, the cathode, current flow, the ½ cell
in which oxidation occurs, and the ½ cell where reduction occur. Finally
write the net reaction that occurs in the cell, the expected potential of the cell,
and make a line diagram of the cell.
Since drawing are hard to do on the computer, I will try to describe
Current flows from left to right
Right side
Positive pole
cathode
oxidation reaction
uses Yttrium ½ reaction
electrode is made of Y
Y+3 is in solution
Left side
Negative pole
anode
reduction reaction
Indium ½ reaction
Since ions can’t make an electrode you need an inert Pt electrode
In+ and In+3 in solution
Line diagram:
Y|Y+3||In+,In+3|Pt
Net reaction:
3In+3(aq) + 2Y(s) W3In+(aq) + 2Y+3(aq)
Potential = -.44 -(-2.38) = 1.94V
4
5.(10 points) Diagram how one of the common reference electrodes is made.
Give the potential of this electrode.
I was looking for diagram like the left half of figure 14-6 for the
Standard Hydrogen Electrode with a potential of 0.00
-orA diagram like figure 14-9 for the silver-silver chloride electrode with a
potential of -.222 V if filled with 1M KCl or +.197 for saturated KCl
-orA diagram like figure 14.10 for the Saturated calomel electrode with a
potential of +.268 if it is filled with 1M KCl or +.241 if filled with saturated
KCl.
6. (10 points) The glass pH electrode, the solid state ion selective electrode,
and the liquid based ion selective electrode all create a membrane potential
by selectively allowing only certain ions to pass through a boundary. (A.)
Explain how this creates a membrane potential, and (B.) Explain how the ion
selective boundary is created for each of these different types of electrodes.
A. Whenever ions are selectively passed through a membrane, a charge
imbalance builds up between the two sides of the membrane, and this results
in the ‘membrane potential’. For example say a membrane passed only +
ions, then + ions move to one side of the membrane leaving their - charged
counter ions behind. Thus one side of the membrane now is + while the other
is negative.
B. In a pH electrode the membrane is actually a very thin glass membrane.
H+ ions are selectively bound on one side of the membrane. This displaces
some Na+ ions on the interior of the glass, which move toward the other side
of the glass, in turn displacing H+ ion on the interior surface of the glass. The
net effect is similar to selectively passing H+ across the membrane surface.
In a solid-state ion-selective electrode the boundary is make of a crystal
that is insoluble the solution being measured, but one of the ions of the
crystal is the ion you wish to measure in the solution. Since this ion is in the
solution, it can enter into the crystal matrix, and can be selectively passed
through the membrane, giving you a membrane potential.
In a liquid-based ion-selective electrode, there is a hydrophobic
membrane that contains the ion you want to measure complexed with a
hydrophobic complexing agent. Since Ions cannot cross a hydrophobic
membrane no other ions can cross, but the target ion, because it can bine to
the hydrophobic complexing agent can cross, so again an ion-selective flow
can be established.
5
Take Home Portion
Do either option A or option B.
Option A
Problem 1 (20 points)
Zn(IO3)2 is marginally soluble, with a Ksp of 3.9x10-6
A. Ignoring any ionic strength effects, calculate the concentration of
Zn+2(aq) and IO3-(aq) in a saturated solution of Zn(IO3)2
B. Based on the concentration of Zn+2 and IO3- in part A, calculate the
ionic strength of the solution.
C. Based on the above ionic strength, calculate the activity coefficients
for Zn+2 and IO3- in this solution, and use these numbers to calculate
new concentrations for Zn+2 and IO3- in the saturated solution.
D. Iterate through this process as necessary to find the true
concentration of Zn+2 and IO3saturated Zn(IO3)2 solution, including activity effects.
A. Solubility of Zn(IO3)2
Ksp = 3.9x10-6
[Zn]=X; [IO3-] = 2X
3.9x10-6 = [Zn][IO3-]2
=X(2X)2
=4X3
X = cube root (3.9x10-6/4)
= 9.92x10-3 = [Zn+2] , 1.983x10-2 = [IO3-]
B. Calculate the ionic strength of this solution
(22×9.92x10-3 + -12× 1.983x10-2 )/2
2.974x10-2 M
6
C. Calculate activity coefficients based on above ionic strength
Zn2+
IO3-
Now
Ksp = 3.9x10-6
[Zn]=X; [IO3-] = 2X
3.9x10-6 = ((Zn[Zn])×((IO3[IO3-])2
=(.546 X)×(2(.85)X)2
=1.578X3
X = cube root (3.9x10-6/1.578)
= 1.35x10-2 = [Zn+2] , 2.704x10-2 = [IO3-]
7
So our concentrations have changed significantly, so the ionic
strength has also changed, and we have to start over. At this point I set
up an excel spreadsheet to do all the above calculations. After I got an
answer for one run, I would then copy the calculations over, and plug the
new numbers to it and try again.
Summary of iterations:
Initial
+2
[Zn ]
.009916
[IO3 ]
.019832
Ionic Strength .0297
(Zn
.546
(IO3
.851
2
.01351
.02702
.04053
.507
.833
3
.0140
.0281
.0421
.503
.831
4
.0141
.0282
.0423
.503
.831
Problem 2 (20 points)
I am going to titrate a 30.00 ml solution of 0.00500M Sn2+ in 1M HCl
with .0200 M Ce4+. (Sn+2 will be in the beaker, the Ce4+ will be in the buret.)
The products of this oxidation/reduction reaction are Sn4+ and Ce3+.
Calculate the potentials you would observe in this titration (vs a Standard
Hydrogen electrode) at the following points: 1.) initial, 2.) 5 ml before the
equivalence point, 3.) at the equivalence point, 4.) 5 ml after the equivalence
point.
½ reactions:
Sn+4 + 2e- 6Sn+2 Eo = .139
Ce+4 + 1e- 6Ce+3 Eo = 1.47
Net reaction:
2 Ce+4 + Sn+2 6Sn+4 + 2 Ce+3
Location of equivalence point
Mole of Sn+2 to titrate = 30 ×.005 = .15 mmoles
.15 mmoles of Sn+2 × (2 mole Ce+4/1mole Sn+2) = .3 mmoles Ce+4
M = mole/volume
.02 = .3/X
X=.3/.02 = 15 ml
8
The points I have to calculate then are
0, 10, 15, and 20 ml
Here is one of my mistakes. You actually can’t calculate the
potential of the initial point, because you only have Sn+2 in the solution and
no Sn+4 so the Nernst equation blows up. For my answer her I will substitute
.01ml of titrant for the 0 point.
.01 ml titrant
Amounts
Sn+2 = 30ml×.005 = .15mmole
Ce+4 = .01ml ×.02 =.0002 mmole
Volume = 30.00 + .01 = 30.01
Reaction table
2 Ce+4
+
Sn+2 6
Initial
.0002
.15
Reaction -.0002
-.0001
Net
0
.1499
Sn+4 +
2 Ce+3
+.0001
+.0002
.0002
.0001
E (Using Sn ½ reaction)
E = .139 - .05916/2 log [Sn+2]/[Sn+4]
=.139-.02958×log[(.1499/30.01)/.0001/30.01)]
=.045V
10 ml titrant
Amounts
Sn+2 = 30ml×.005 = .15mmole
Ce+4 = 10ml ×.02 =.2 mmole
Volume = 30.00 + 10 = 40
Reaction table
2 Ce+4
Initial
.2
Reaction -.2
Net
0
+
-.1
.05
Sn+2 6
.15
+.1
.1
E (Using Sn ½ reaction)
E = .139 - .05916/2 log [Sn+2]/[Sn+4]
=.139-.02958×log[(.05/40)/(.1/40)]
=.148V
Sn+4 +
+.2
.2
2 Ce+3
9
15 ml
Equivalence point
E = [2(.139) + 1(1.47)] /3
= .583V
20 ml
Amounts
Sn+2 = 30ml×.005 = .15mmole
Ce+4 = 20ml ×.02 =.4 mmole
Volume = 30.00 + 20 = 50
Reaction table
2 Ce+4
Initial
.4
Reaction -.3
Net
.1
Sn+2 6
.15
-.15
+.15
0
.15
+
E (Using Ce ½ reaction)
E = 1.47 - .05916/1 log [Ce+3]/[Ce+4]
=1.47-.05916×log[(.3/50)/(.1/50)]
=1.44V
Sn+4 +
+.3
.3
2 Ce+3
10
Option B
In lab we titrated my water sample with EDTA, and had some difficulty
seeing where the endpoint was. Perhaps we could have used electrodes to
follow the titration instead of using an indicator.
Problem 1 (10 points) Propose one or more electrodes you could build in the
lab to follow the EDTA titration of Mg2+ and Ca2+.
Problem 2 (30 points) Assume you had 10 mls of solution that was 6mM Ca2+
and 3 mM Mg2+ buffered at pH 10, and you were titrating this solution with 3
mM EDTA. Calculate the potential your electrode(s) would give you at 0,
10, 20 and 30 ml of EDTA added.
Nobody tried this option, so I won’t give you a key