Name:
Chem 332
Analytical Chemistry
Exam I
Show all work for partial credit
1. Make a rough sketch of the titration curve of 10 mL of .1M oxalic acid (H2C2O4) when
it is titrated with .1M KOH. Oxalic acid has pKa’s of 1.28 and 4.27. What are the
principal species in the solution after 1, 10, 14, and 19 mLs of titrant have been added
to the solution. (Do not try to calculate exact pH’s as any point in this titration curve!)
Titration curve:
Hard to do with a word processor
Should be a titration curve with 2 buffer regions
The first buffer region is centered at 5 mls of KOH and pH of 1.28
The second buffer region is centered at 15 mLs and pH =4.27
The inflection point between the two buffers is at 10 mLs and pH 2.8
Principal species:
1 mL H2C2O4
10 mL HC2O414 mL HC2O419 mL C2O422. What is the ionic strength of a solution that is .5M NaCl, .25M CaCl2, and .1M
Li3PO4?
[Na+]=.5, [Cl-] = .5 + 2(.25) = 1.0, [Ca2+] = .25, [Li+] = 3(.1) = .3, [PO43-] = .1
Ionic strength
= [.5(1)2 + 1(-1)2 + .25(2)2 + .3(1)2 + .1(-3)2]/2
=[.5+1+1+.3+.9]/2
=1.85 M
3A.The Extended Debye-Hückel equation is:
What is the activity coefficient for Fe3+ if its ion size it 900 pm and the ionic strength of
the solution is 0.02M?
3B. What is the activity coefficient of Th4+ at an ionic strength of .07, if it has an activity
coefficient of .1 at an ionic strength of .05, and an activity coefficient of .065 when the
ionic strength is .1?
X(ì) Y(ã)
.05
.1
.1
.065
.07
?
I keep getting fouled up with the book method of interpolation, so I am going to do this
with the equation for a line, Y=mX+b. The slope of a line is ÄY/ÄX = (.065-.1)/(.1-.05)
= -.7
Once we have a slope we can plug the X and Y of either of the two points into
Y=mX +b to find b. I’ll use the first point. .1 = -.7(.05) +b, .1=-.035+b; b=.135. Now I
have m and b I can use the equation of the line Y = -.7(.07) + .135; Y =-.049 + .135;
Y=.086
4. Define the following terms:
auxilliary complexing agent - A chemical that forms a complex with the metal ion
that prevents the formation of metal hydroxides that can precipitate as a solid.
Kf’ = áY-4Kf = The formation constant multiplied by a factor that accounts for pH
Masking agent A chemical that is added to form a strong complex with one metal
ion in a mixture of metal ion that prevents that metal from being titrated with EDTA.
Thus allowing the other metal to be titrated as if the first metal was ‘masked’ so it would
not appear in the titration.
Coordinate covalent bond The bond found in metal ion complexes. It is not as
strong as a regular covalent bond because both of the electrons in the bond come from
the anion so the anion can be replaced by another anion with its own pair of electrons.
A metal that blocks an indicator A metal that binds so strongly to the indicator
that it cannot be displaced by EDTA, so no endpoint is seen in the titration of the metal
with EDTA.
5 .When we did polyprotic acids we said that when the Ka of the acids were far enough
apart, we could treat each ionization independently, and this made all the calculations
easier. Then, in a later chapter, I gave you the tools (the systematic approach to
equilibrium) to solve even a difficult problem when the Ka’s are close together. Try this.
What is the pH of .01M Thiosulfuric acid (H2S2O3) with Ka’s of .3 and .03?
Actually, don’t try to solve this right now.
A. Write down all the species you will have in this solution
H2S2O3, HS2O3-,S2O32-, H+, OH-, H2O
B. Write down the charge balance equation for this solution
[H+] = [HS2O3-] + 2[S2O32-] + [OH-]
C. Write down the mass balance equation for this solution
.01 = [H2S2O3]+[HS2O3-] + [S2O32-]
D. Write down all equilibrium equations that apply to this solution.
E. Do you have enough equations to solve the system?
6 unknowns 5 equations, now you do not have enough equations to solve.
F. Can you make any assumptions to simplify the system or eliminate some
species from the problem?
Assume [H2O]=55.5M Now have 5 equations and 5 unknowns and can solve.
Can also assume that this is an acid solution so [OH-] in the charge balance
equation is so small that it can be neglected.
6. Given that Cr3+ has a log Kf of 23.4, and the áY-4 for EDTA is 2.1x10-11 at pH 3, is the
titration of Cr3+ feasible at pH 3?
Log Kf = 23.4; Kf = 1023.4, Kf = 2.5x1023
Kf’ = 2.1x10-11 × 2.5x1023 = 5.3x1012
Kf’ must be >108 to be feasible, so this titration is feasible.
7. If we have 25 mLs of a .005M metal solution, and the Kf’ for this metal and EDTA is
5x1011.
What is the p[Metal] after the following amounts of .003M EDTA have been
added to the metal solution.
A. 0 mLs of EDTA added
p[Metal] = -log(.005) = 2.3
B. 25mLs of EDTA added
25 mLs x .005 M metal = .125 mmoles metal
25 mLs x .003M EDTA = .075 mmoles
I
C
E
ICE table
Metal
.125
-.075
.05
EDTA
.075
-.075
0
MetalAEDTA
0
+.075
+.075
p[Metal] = -log (.05mmoles/50 mLs) = 3
C. 50 mLs of EDTA added
25 mLs x .005 M metal = .125 mmoles metal
50 mLs x .003M EDTA = .150 mmoles
I
C
E
ICE table
Metal
.125
-.125
0
EDTA
.150
-.125
.025
MetalAEDTA
0
+.125
+.125
Since [Metal] = 0, the equilibrium will shift to fill this in, so
E
+X
.025+X
.125-X
Kf’ = [MetalAEDTA]/[Metal][EDTA]
Lets shift to molar concentration so we can plug into the above equation
[Metal] = X
[EDTA] = .025/(25+50) +X = 3.33x10-4 +X
[MetalAEDTA] = .125/(25+50) -X = .00167-X
But Kf’ is 5x1011 so we can assume the +X and -X terms are negligible so:
And X = 1x10-11 is small enough that our assumptions were correct.
Take home portion
1 Ba(OH)2 has a Ksp of 3x10-4. What is the solubility in deionized water? What is the
solubility in .1M NaCl (include activity effects!)
Deionized water - no activity
Ba(OH)2 WBa2+ + 2OHKsp = 3x10-4=[Ba2+] + [OH-]2
X=[Ba2+]; [OH-]=2X
3x10-4=[X] × [2X]2 =X4X2
=4X3
X=(3x10-.4/4).33333333
.0422 =[Ba2+]; [OH-]=.0844
in .1M NaCl ì=.1. You can either calculate or look up the activity coefficients of
Ba and OH- @ ì=.1. Looking up in the table is easiest!
ãBa = .38 ãOH= .76, but I gave you 2 bonus point for calculating it!
2+
Our KSP expression not becomes:
3x10-4=.38[X] × [2(.76X]2 =.888X3
X=(3x10-.4/.888).33333333
X=.0696
.0696 =[Ba2+]; [OH-]=.139
2.Cis-Butenediotic acid (HOOC-CH=CH-COOH) has two acid functional groups with
pKa’s of 1.92 and 6.27. Calculate the pH of .001M NaOOC-CH=CH-COOH (the
species found at the first equivalence point in the titration of this acid) using the exact
equation and the quick and dirty calculation. If the answers are different explain why.
Quick and dirty:
pH = (pK1 + pK2)/2 = (1.92+6.27)/2 = 4.095
Long expression:
K1 =.0120
K2 =5.37x10-7
Which is much different than the quick and dirty. The answers are different
because K1 is larger than the concentration. In the quick and dirty equation we assume
that K1 is < the concentration
3. Returning to the Thiosulfuric acid problem I gave you in the in class portion of the
test. Can you determine the pH of .01M Thiosulfuric acid? I haven’t actually tried this
yet, so I don’t know if it can be solved. So show me how far you can get. Come up with
a set of equations to solve the system, then start combining the equations to eliminate
species. How far can you get? Can you get to a single equation with a single
unknown?
Here are the potential species we need to solve for:
H2S2O3, HS2O3-,S2O32-, H+, OH-, H2O
Of these we will assume that H2O is a constant, at 55.5 M so we only have to solve for
the 5 concentrations:
[H2S2O3] [HS2O3-], [S2O32-], [ H+], & [OH-]
I can come up with 5 equations
[H+] = [HS2O3-] + 2[S2O32-] + [OH-]
EQN 1
2.01 = [H2S2O3]+[HS2O3 ] + [S2O3 ]
EQN 2
EQN 3
EQN 4
1x10-14 = [H+][OH-]
EQN 5
Here is the algebra approach I took. You will see that I eventually got to a single (very
hairy) equation with a single unknown, but the solver function on my calculator could
not solve it, so I don’t know if my algebra was wrong, or if I plugged it into my calculator
wrong. If you can see an error in my math please correct me!
OH- small so EQN 1 changes to
[H+] = [HS2O3-] + 2[S2O32-]
EQN 1'
Now substitute the above equation for [H+] into both equilibrium equations to eliminate
[H+] from our set of equations:
EQN 3'
EQN 4'
Now play with 4' to try to get [HS2O3-] all by itself
I’ll call the last equation EQN4"
Now go back to equation 2 and rearrange it to get [H2S2O3] all by itself
.01 = [H2S2O3]+[HS2O3-] + [S2O32-]
.01 - [HS2O3-] -[S2O32-] =[H2S2O3]
Eqn 2'
And plug this back into equation 3' to eliminate [H2S2O3] from the set of equations
EQN 3"
So now we can substitute eqn 4" into equation 3'’ to get a single equation with a single
unknown, that maybe the equation solver can handle.
0=(((2x2/(.03-x))+2x)×(2x2/(.03-x)))/(.01-(2x2/(.03-x))-x)-.3
Solver did not solve for X,
it into my calculator wrong.
So I don’t know if the expression was wrong, or if I plugged
So let’s try another approach, the successive approximations method. Here are the two equations
I want to use:
First time through, assume Y=0
.3=X2/(.01-X)
0=[X2/((.01-X)]-.3
The solver can give you 2 roots: -.309,.009687
I’ll use the + root on the next equation:
.03=(.009687+2Y)Y/.009687-Y
Again two roots: -.02553,.005691
0=[(.009687+2Y)Y/.009687-Y]-.03
Now with the subsequent iterations I will use the full equations:
.3=(X+2(.005691)X/(.01-X-.005691)
-.3155,.004098
.03=(.004098+2X)X/.004098
-.00893, .006882
Third time
.3=(X+2(.006882)X/(.01-X-.006882)
-.3169,.003141
.03=(.003141+2X)X/.003141
.006124
Third time
.3=(X+2(.006124)X/(.01-X-.006124)
-.3159,.003681
.03=(.003681+2X)X/.003681
-.00841, .006570
Fourth time
.3=(X+2(.006570)X/(.01-X-.006570)
-.3164,.003252
.03=(.003252+2X)X/.003252
-.00784, .006218
Fifth time
.3=(X+2(.006218)X/(.01-X-.006218)
-.3160,.003590
.03=(.003590+2X)X/.003590
-.00829, .006495
0=(X+2(.005691)X/(.01-X-.005691))-.3
0=((.004098+2X)X/.004098)-.03
0=(X+2(.006882)X/(.01-X-.006882))-.3
0=((.003141+2X)X/.003141)-.03
0=(X+2(.006124)X/(.01-X-.006124))-.3
0=(.003681+2X)X/.003681)-.03
0=(X+2(.006570)X/(.01-X-.006570))-.3
0=(.003252+2X)X/.003252)-.03
0=(X+2(.006218)X/(.01-X-.006218))-.3
0=(.003590+2X)X/.003590)-.03
Sixth time
.3=(X+2(.006495)X/(.01-X-.006495)
-.3163,.003324
.03=(.003324+2X)X/.003324
-.00794, .006279
0=(X+2(.006495)X/(.01-X-.006495))-.3
0=(.003324+2X)X/.003324)-.03
The successive approximations haven’t settled on an exact answer, but you can see the method is
working, and [HS2O3-] will be around .0034 and [S2O32-] will be around .0064. The rest you can
get by simple substitution.
[H2S2O3]= .01-.0034-.0064 = .0002
[H+]=.0034+2(.0064) = .0162, pH = 1.79, pOH = 12.01
Jake, Jayce, Taylor, and Sam came up with a better approach.
[H+] = [HS2O3-] + 2[S2O32-] + [OH-]
EQN 1
.01 = [H2S2O3]+[HS2O3-] + [S2O32-]
EQN 2
EQN 3
EQN 4
1x10-14 = [H+][OH-]
EQN 5
Assume that [OH-] in equation 1 is small and can be ignored
AND
Assume [H2S2O3] is small and can be ignored in equation 2. Since H2S2O3 is a weak
acid, we usually cannot make that assumption, but since the second ionization will push
the ionization along, lets give it a try:
[H+] = [HS2O3-] + 2[S2O32-]
EQN 1'
.01 = [HS2O3-] + [S2O32-]
EQN 2'
[HS2O3-]=.01-[S2O32-]
EQN 2"
Put this into equation 1
[H+] = .01-[S2O32-] + 2[S2O32-]
[H+] = .01+[S2O32-]
[S2O32-]=[H+]-.01
EQN A
And plugging this into Eqn 2"
[HS2O3-]=.01-([H+]-.01)
[HS2O3-]=.02-[H+]
EQN B
Now plug Eqn A and B into equation 4
Which can be solved with the quadratic or the solver
[H+]=.01646 pH=1.78 which you can see is pretty darn close! And a lot simpler than my
successive approximation!