Name:_____________
Chemistry 332
Second Hour Exam
1A (12 points) Do either choice 1 or choice 2, If you do both I will give you
the higher of the two scores
Choice 1. - I have a solution that is .36M Malonic Acid. This is a diprotic
acid with pKa’s of 2.85 and 5.70. I also have a solution of .5M KOH. What
is the pH in the following mixtures:
5 ml of malonic acid and 1 ml of KOH
5 ml x .36M = 1.8 mmole MalH2
1 ml x .5M = .5 mmole KOH
MalH2 + KOH 6MalH + H2O
1.8
.5
0
0
-.5
-.5 +.5
1.3
0
.5
pH = pKa1 + log (MalH/MalH2)
= 2.85 + log (.5/1.3)
=2.435
5 ml of malonic acid and 5 ml of KOH
5 ml x .36M = 1.8 mmole MalH2
5 ml x .5M = 2.5 mmole KOH
MalH2 + KOH 6MalH + H2O
1.8
2.5 0
0
-1.8
-1.8 +1.8
1.8
0
.7
1.8
MalH + KOH 6 Mal + H2O
1.8
.7
0
0
-.7
-.7 +.7
0
1.1
0
.7
0
pH = pKa2 + log (Mal/MalH)
= 5.7 + log (.7/1.1)
= 5.504
5 ml of malonic acid and 10 ml of KOH
5 ml x .36M = 1.8 mmole MalH2
10 ml x .5M = 5 mmole KOH
MalH2 + KOH 6MalH + H2O
1.8
5
0
0
-1.8
-1.8 +1.8
1.8
0
3.2 1.8
2
MalH + KOH 6 Mal + H2O
1.8
3.2 0
0
-1.8
-1.8 +1.8
0
0
1.4 1.8
0
pH determined by excess KOH
1.4 mmole/15 mls = .0933M
pOH = -log(.0933) = 1.03; pH = 14-1.03 = 12.97
Choice 2.
What is the pH of .36M Malonic acid (HO2CCH2CO2H)?
Ka = X2/(.36-X)
Ka = 10-2.85 = 1.413x10-3
1.413x10-3= X2/(.36-X)
Assume .36 >>X
1.413x10-3= X2/(.36) ; X = sqrt(1.413x10-3(.36)) = 2.25x10-2
.0225 ~ 6% of .36, so error is a bit big, try a successive
approximation
1.413x10-3= X2/(.36-.0225); X = sqrt((1.413x10-3(.36-.0225)
X = 2.18x10-2
This should be pretty close so pH = -log(2.18x10-2)
= 1.66
What is the pH of .36M Sodium Malonate (NaO2CCH2CO2H)?
Q&D pH =( pKa1 + pKa2)/2 = (2.85+5.7)/2 = 4.275
What is the pH of .36M Disodium Malonate (NaO2CCH2CO2Na)
Using base equation
KB1 = 10-14/10-5.7 = 10-8.3 = 5.01x10-9
5.01x10-9 = X2/(.36-X)
K small so .36-X ~ .36
5.01x10-9 = X2/.36; X=sqrt(5.01x10-9 x .36) = 4.25x10-5
4.25x10-5 = [OH-]; pOH = -log(4.25x10-5) = 4.37
pH = 14-4.37 = 9.63
3
1B. (6 points) What are the principle species of Malonic acid at
pH 2 H2 Mal (A little below pKa1, so H2Mal dominate)
pH 4 H Mal
Entirely in Hmal form at pH = (pKa1 + pKa2)/2
or pH = 4.27, this is just a littel off, so same form
pH 6 Mal
A little past pKa2 so into fully deprotonated form
2. (10 points) What is the ionic strength of a solution of .1M MgCl2 and .2M
Na2SO4?
[.1(22) + .2(-12) + .4(12) + .2(-2)2]/2
= .9M
3. (5 points) K+, Ca2+, and Sc3+ could be expected to have the same hydrated
radius. Would they have the same activity coefficient? Why or why not?
Which ion would have the smallest coefficient?
The activity coefficient gets smaller as the charge on the ion increases,
so you would expect Sc+3 to have the smallest activity coefficient.
4. (5 points) Be2+, Mg2+ and Ca2+ have the same net charge. Would you
expect them to have the same activity coefficient? Why or why not? Which
ion would have the smallest coefficient?
The activity coefficient gets smaller as the hydrated radius gets
smaller, so you would expect Be+2 with the smallest ionic radius to have the
smallest activity coefficient. However, I would also accept the argument that
Be2+ would have the smaller ionic radius, so this would make more water
bind, so it would actually have the largest hydrated radius, so in should have
activity coefficient closest to 1.
4
5. (10 points) The association between V2+ and EDTA has a log Kf of 12.7.
Use the following table to determine the minimum pH at which EDTA can
be used to titrate V2+
pH
"Y-4
8
4.2x10-3
7
3.8x10-4
6
1.8x10-5
5
2.9x10-7
To have a good titration Keff > 1x10-8
Keff = " * Kf
Kf = 1012.7 = 5.01x1012
" x 5.01x1012$108
108/5.01x1012 $"
2x10-4 $"
pH 7, with " = 3.8x10-4 will work, but not any lower.
5
6. (12 points) Use the following two reactions to build a Galvanic Cell.
Make a diagram of this cell with the following points labeled: + pole, - pole,
anode, cathode, flow of electrons, salt bridge, cell where oxidation occurs,
cell where reduction occurs, chemical content of the right hand cell,
chemical contents of the left hand cell. Also make a line diagram of the cell.
What is the potential of this cell under standard state conditions. What is
the potential of the cell when all aqueous chemicals are at 1x10-4M
concentrations?
Reactions to use
ClO3- + 6H+ + 5e- 6 ½ Cl2(g) + 3H2O
Sr2+ + 2e- 6Sr
E0 = 1.458
E0 = -2.889
Left
Right
+
anode
cathode
electrons flow from left to right
oxidation
reduction
2+
Sr + 2e 6Sr
ClO3- + 6H+ + 5e- 6 ½ Cl2(g) + 3H2O
Sr*Sr2+2ClO3-,H+,H2O,Cl2(g)*Pt
E0 = 1.458 - (-2.889) = 4.347
Non-standard conditions:
EL = -2.889 -.059/2log(1/1x10-4)
= -2.889 - .0295(4)
=-2.889-.118 = -3.007
ER = 1.458 - .059/5 log (1/[(1x10-4)(1x10-4)6]
=1.458 -.0118 log(1x10+28)
=1.458 -.0118(28)
=1.458 - .330
=1.128
E = ER - EL = 1.128 -(-3.007) = 4.135
6
Take home portion - Use a spreadsheet program if you wish.
1. (25 points) In the first question of the in class portion of the test I asked
you the pH as a for a couple of mixtures of .36M Malonic Acid and .5M
KOH. Now let’s do it right. Calculate a titration curve for the titration of
.36 M Malonic Acid with .5M KOH. Include the following points in your
curve:
A. Inital pH.
B. pH and volume of KOH 1/3 of the way to the first equivalence
point.
C. pH and volume of KOH at the first equivalence point.
D. pH and volume of KOH 3/4 of the way between the first and
second equivalence points
E. pH and volume at the second equivalence point.
F. pH and volume 5 mls after the second equivalence point.
G. Based on these calculations suggest an indicator to be used to find
the first equivalence point.
H. Based on these calculations suggest an indicator to be used to find
the second equivalence point.
I forgot to give you a volume, so you can pick any of Malonic yo want,
but just to match the in-class portion I will go with 5 mls of Malonic.
1st equivalence point
5(.36) = X(.5)
X = 5(.36)/.5 = 3.6 ml
nd
2 equivalence point = 2(3.6) = 7.2
A. Initial pH
Ka = X2/(.36-X)
Ka = 10-2.85 = 1.413x10-3
1.413x10-3= X2/(.36-X)
Assume .36 >>X
1.413x10-3= X2/(.36) ; X = sqrt(1.413x10-3(.36)) = 2.25x10-2
.0225 ~ 6% of .36, so error is a bit big, try a successive
approximation
1.413x10-3= X2/(.36-.0225); X = sqrt((1.413x10-3(.36-.0225)
X = 2.18x10-2
Using a spreadsheet to interate this calculation
st
1 approx X = 2.25x10-2
7
2nd appox X = 2.18x10-2
3rd approx X=2.18x10-2; pH = 1.66
B . 1/3 way to first EQ. PT
.333(3.6) = 1.2 ml
moles H2 Mal = 5(.36) = 1.8 mmole
moles KOH = 1.2(.5) = .6 mmoles
moles H2 Mal remaining = 1.8-.6 = 1.2
mole H Mal = .6
pH = pKa1 + log (H Mal/ H2Mal)
= 2.85 + log (.6/1.2)
= 2.549
st
C. 1 equivalence point
The Q & D equation
pH =( pKa1 + pKa2)/2 = (2.85+5.7)/2 = 4.275
The big equation
[H+ ]= sqrt((K1K2[H Mal] + K1Kw)/(K1+[H Mal])
K1 = 1.41x10-3
K2 = 2.00x10-6
[H Mal] = 1.8 mmole/(5+3.6 ml) = .209M
=sqrt((5.89x10-10+1.41x10-17)/.2104)
= 5.29x10-5 ; pH = 4.276
Not a heck of a lot of difference for all the work!
D. 3/4 of the way between the 1st and 2nd Eq Pt
.75(3.6) = 2.7
Vol = 3.6 + 2.7 = 6.3 ml
Moles H2 Mal = 1.8 mmole
Moles KOH = 6.3(.5) = 3.15 mmole
use up 1.8 mmole of base converting H2 Mal to H Mal
so 3.15 - 1.8 = 1.35 mmoles KOH remain to react with H Mal
H Mal remaining = 1.8 mmole -1.35 = .45 mmole H Mal
Mal = 1.35 mmole
pH = pKa2 + log (Mal/ H Mal)
= 5.70 + log (1.35/.45)
8
= 6.18
9
E. Second equivalence point
Vol = 7.2 ml of base
Using base equation
KB1 = 10-14/10-5.7 = 10-8.3 = 5.01x10-9
[Mal] = 1.8 mmole / (5 + 7.2 ml) =.148M
5.01x10-9 = X2/(.148-X)
K small so .148-X ~ .148
5.01x10-9 = X2/.148; X=sqrt(5.01x10-9 x .148) = 2.72x10-5
2.72x10-5 = [OH-]; pOH = -log(2.72x10-5) = 4.56
pH = 14-4.37 = 9.44
F. 5 ml after
Vol = 7.2 + 5 = 12.2 ml
Excess base = 5(.5) = 2.5 mmole
volume = 12.2 + 5 = 17.5
molarity of excess base = 2.5 mmol/17.5 ml = .143M
pOH = -log(.143) = .845; pH=13.155
G. Indicator for 1st Eq Pt at pH 4.275 is Ethyl Orange pKa 4.1
H. Indicator for 2nd Eq Pt at pH 9.44 Thymolphthalein
2. (15 points)
The Ksp of CaSO4 is 2.4x10-5
A. Calculate the solubility of this salt, assuming that the activity
coefficient of both ions is 1.
Ksp = 2.4x10-5 = [Ca2+][SO42-]
[Ca2+]=[SO42-] = X
2.4x10-5 = X2
X=sqrt(2.4x10-5) = 4.899x10-3
B. Now calculate the true solubility of this salt by including activity
terms. You may have to iterate through the problem a couple of
times to find the correct answer by successive approximation.
Based on the above concentrations, the ionic strength of the solution
has become [4.899x10-3(+2)2 + 4.899x10-3(-2)2 ]/2 = 1.96x10-2
Now you have to plut this ionic strength in to the extended debye-Huckel
equation to find ( for each ion
10
For Ca2+ "=600;
Log( = -.224
(=10-.224
( = .597
For SO42- "=400
Log ( = -.241
( = 10-.241
( = .574
Ksp = 2.4x10-5 = X(.597)X(.574)
X = sqrt(2.4x10-5/(.597)(.574)
X = 8.37x10-3
Now plugging that in for multiple iterations
X
u
4.90E-03
0.008369
0.009503
0.009822
0.009909
0.009932
0.009939
0.00994
0.009941
1.96E-02
3.35E-02
3.80E-02
3.93E-02
3.96E-02
3.97E-02
3.98E-02
3.98E-02
log g Ca log g SO4 g Ca
g SO4
-2.24E-01 -2.41E-01 5.97E-01 5.74E-01
-2.74E-01 -3.01E-01 5.32E-01 5.00E-01
-2.87E-01 -3.17E-01 5.16E-01 4.82E-01
-2.91E-01 -3.21E-01 5.12E-01 4.78E-01
-2.92E-01 -3.22E-01 5.11E-01 4.76E-01
-2.92E-01 -3.22E-01 5.10E-01 4.76E-01
-2.92E-01 -3.22E-01 5.10E-01 4.76E-01
-2.92E-01 -3.22E-01 5.10E-01 4.76E-01