CHEM*2400/2480
Summer 2004
Assignment 5
ANSWERS
1. (a) The only ions will be lanthanum, fluoride, hydroxide, and hydronium. We find
3 6 La 3 + @ + 6 H + @ = 6 F - @ + 6OH - @
This is a charge balance equation so that the charge on the ion comes in as the coefficient
in front of each species (the 3 with the La3+ ion).
(b)
More ions can be present and we would now write
3 6 La 3 + @ + 2 6 LaF 2 + @ + 6 LaF2+ @ + 6 H + @ = 6 F - @ + 6OH - @
2. Be sure to account for all forms of the conjugate base of the molecule. And watch the
stoichiometry of the charge. The doubly charged anion form will contribute twice the
charge to the balance of the equation.
6 H + @ = 6 HOOCC 2 H 2 COO - @ + 2 6- OOCC 2 H 2 COO - @ + 6OH - @
3. (a) Now we know that the total lanthanum in solution will be 0.20 mol/liter. This is
because there is 1 mole of La per mol of LaF3. We need simply identify all of the forms
in which it might hide.
0.20 M = 6 La 3 + @ + 6 LaF 2 + @ + 6 LaF +2 @
The only forms lanthanum can be found in have it present with the stoichiometry of 1, so
that the coefficient in each case is just 1. Recall that in the mass balance equation we are
balancing amount of material rather than charge so it is the subscripts on the ions (rather
than the superscripts) that show up as coefficients. Also, remember that we are writing
an equation for lanthanum and not fluoride; that comes next.
(b) Same thing, except that the total F concentrations will be 3 x 0.20 = 0.60 M
because the fluoride stoichiometry is 3 in the original molecule. This time we will have
to watch out for the stoichiometry of the fluoride in each ion and account for it in the
coefficients of the equation.
0.60 M = 6 F - @ + 6 LaF 2 + @ + 2 6 LaF +2 @
4. We cannot write a mass balance involving the hydronium ion because we don’t know
how much comes from water dissociation. But we can write one for the maleate ion in its
various forms.
0.20 M = 6 HOOCC 2 H 2 COOH @ + 6 HOOCC 2 H 2 COO - @ + 6- OOCC 2 H 2 COO - @
Reminder: At this point in the process we do not know how the 0.20 M is divided up
between the three components but do know that their sum must equal 0.20 M.
5. For this mass balance equation, we do not know the absolute formal concentration, but
we do know the way in which the thorium species must be related to the fluoride species:
the total thorium must be one-quarter the total fluoride. This is because of the
stoichiometry of the original molecule.
= 14 6F - @ total
6 Th 4 + @ + 6 ThF22 + @ + 2 6 Th 2 F53 + @ + 6 ThF 3 + @ =
6 Th 4 + @
total
1 #6 F - @ + 2 6 ThF 2 + @ + 3 6 Th F5 + @ + 6 ThF 3 + @2
2
3
4
6 Th 4 + @ = 1 6 F - @ - 1 6 ThF 22 + @ - 5 6 Th 2 F 53 + @ - 3 6 ThF 3 + @
4
2
4
4
6. We need to write down the pertinent equations. There will be four equilibrium equations.
They are
AgCN ^sh ? Ag + + CN K sp1 = 7 Ag + A 6CN - @ = 2.20 # 10 - 16
Zn ^CN h 2 ^sh ? Zn 2 + + 2 CN -
2
K sp2 = 6 Zn 2 + @ 6CN - @ = 3.03 # 10 - 16
K w = 6 H + @ 6OH - @ = 1.01 # 10 - 14
6 H + @ 6CN - @
HCN ? H + + CN Ka=
= 6.22 # 10 - 10
6 HCN @
We count up the number of unknowns and find that there are six: Ag+, Zn2+, CN-, H+,
OH-, and HCN. We still need two more equations for the problem to be solvable; they
are, of course, the mass balance and the charge balance equations. The charge balance is
quite straightforward; just remember the stoichiometry on the charges.
6 H + @ + 7 Ag + A + 2 6 Zn 2 + @ = 6OH - @ + 6CN - @
The mass balance is a little trickier. I reasoned it out this way. All of the cyanide ion
must come from the dissociation of the two metal salts. Since 2 CN- ions will arise every
time one Zn2+ ion appears, the zinc contribution to the cyanide must be twice the Zn
concentration. Now the only further wrinkle is that some of the cyanide goes into HCN,
but the total cyanide is still the sum of those two quantities. We would have
7 Ag + A + 2 6 Zn 2 + @ = 6CN - @ + 6 HCN @
We now have six equations and six unknowns so we can solve the problem. If you look
at the equilibrium equations, the cyanide term [CN-] is the most common. Therefore, lets
try and substitute for the non-cyanide terms in the mass balance equation to get an
expression mostly in just cyanide. Either would work but with only four terms in the
mass balance rather than five, it will be easier to see if that works O.K.
7 Ag + A + 2 6 Zn 2 + @ = 6CN - @ + 6 HCN @
H 2 O ? H + + OH -
6 H + @ 6CN - @
K sp1
2 K sp2 6
+
2 = CN @ +
Ka
6CN @ 6CN @
Here is an equation in only 2 unknowns. We can easily solve for [H+] in terms of
cyanide. This is rewritten to read
6 H + @ 6CN - @
K sp1
2 K sp2
=
+
- 6CN - @
Ka
6CN - @ 6CN - @ 2
6 H + @ = K a ) K sp1 2 + 2 K sp2 3 - 13
6CN - @
6CN - @
Now return to the charge balance equation. We can substitute the above expression for
[H+] so it is only in terms of [CN-]. The next two terms can be substituted from the Ksp
expressions to make [CN-] the only variable. The only other term is the [OH-]. The Kw
expression can make it an expression in only [H+] but then we can substitute the above
again. This leaves us with only 1 equation in 1 unknown. It can be solved for [CN-]!
6 H + @ + 7 Ag + A + 2 6 Zn 2 + @ = 6 OH - @ + 6CN - @ = K w+ + 6CN - @
6H @
fK a )
K sp1
2 K sp2
K sp1
K sp2
+2
2 +
3 - 13 p +
2 =
6CN - @
6CN - @
6CN - @
6CN - @
Kw
fK a )
K sp1
2 K sp2
2 +
3 - 13 p
6CN - @
6CN - @
+ 6CN - @
If you expand this out, term by term, you will find you get a 6th order polynomial in [CN]. The challenge, of course, is to find the one valid root for the expression. It is time for
the spreadsheet and the brute force method. Bring the right hand side terms to the left
and set the entire expression equal to 0. The challenge is to find the value of cyanide
concentration for which the expression is an equality. We choose a range of cyanide
concentrations (again I like to spread it out in logarithmic space, and convert that to
concentration, then solve this complicated expression to see in what range of
concentrations the sum changes from positive to negative or vice versa. We then
decrease the step size (go from 0.05 to 0.005 in the log(CN-) array) and find the new zero
crossing range. Continue to “zoom” in on the zero crossing until you have enough
significant figures. I think it is important to carry many additional significant figures
until the very end. Round off is always a potentially error-filled exercise, occurring when
subtracting two values that are very close in magnitude. With Excel, I found that, out to
9 significant figures, the correct concentration of cyanide was somewhere between
6.1329080 x 10-6 and 6.1329087 x 10-6 M.
Fixing [CN-] = 6.132908 x 10-6 M, we can find [H+] from the third equation above. Then
Kw gives us the value for [OH-], the Ka expression solves for [HCN], and the two Ksp
expressions solve for the two metal ion concentrations. The final results, rounded to 3 sig
figs are:
[CN-] = 6.13 x 10-6 M
[H+] = 1.01 x 10-9 M
[OH-] = 9.98 x 10-6 M
[HCN] = 9.98 x 10-6 M
[Ag+] = 3.59 x 10-11 M
[Zn2+] = 8.06 x 10-6 M
Note that the hydroxide and the HCN concentration are given as being equal. Does this
make sense? We are saying that whenever a cyanide ion reacts with water, it produces an
hydroxide and an HCN. They would be exactly equal except that water also produces
some hydroxide. But with a cyanide concentration of approaching 1 x 10-5 M, the
dominant source of hydroxide will be from cyanide. Indeed, the amount of hydroxide
coming from water dissociation is only that equal to the amount of hydronium ion which
is around 10-9 M. If we carry the calculation out to more significant figures, we do see
that the hydroxide is slightly greater than the HCN, different in the third decimal place,
exactly as we would predict.
7. This is similar to the last problem except that there are 8 species whose concentration
need to be specified. There are six equilibria that can be specified readily. They are
PO34 - + H 2 O ? HPO24 - + OH HPO 24 - + H 2 O ? H 2 PO -4 + OH H 2 PO -4 + H 2 O ? H 3 PO 4 + OH Pb 2 + + H 2 O ? PbOH + + H +
H 2 O ? H + + OH -
6 HPO 24 - @ 6 OH - @
= 2.37 # 10 - 2
6 PO 34 - @
6 H 2 PO -4 @ 6OH - @
K b2 =
= 1.59 # 10 - 7
6 HPO 24 - @
K b1 =
K b3 =
6 H 3 PO 4@ 6OH - @
= 1.38 # 10 - 12
6 H 2 PO -4@
6 PbOH + @ 6 H + @
= 2.48 # 10 - 8
6 Pb 2 + @
K w = 6 H + @ 6OH - @ = 1.01 # 10 - 14
Kc =
3
2
Pb 3 ^PO 4h 2 ? 3 Pb 2 + + 2 PO 34 K sp = 6 Pb 2 + @ 6 PO 34 - @ = 3.03 # 10 - 44
We still need the mass balance equation. It comes from thinking about what the balance
between the total lead and the total phosphate will be. They can only come from the
solvation of the salt. We see that 3 Pb must enter solution whenever 2 phosphates enter.
Therefore, the total phosphate concentration must be 2/3 of the total lead concentration.
This leads to the mass balance expression.
2 6 Pb@ = 6 PO @
4
total
total
3
2 _6 Pb 2 + @ + 6 PbOH + @i = PO 3 - + HPO 2 - + H PO - + 6 H PO @
3
4
6 4 @ 6
6 2 4@
4 @
3
To completely solve the problem we need to also write down the charge balance
equation. However, because we have stated that in this problem we have fixed the pH at
5.500, the charge balance cannot be written down since there must be additional,
unspecified materials in the solution acting as a pH buffer. But the great thing, is that
while we can’t write down the expression, we do not need to because fixing the pH,
directly specifies the values for 2 of our unknowns, namely [H+] and [OH-]. We now
have 7 equations and only 6 unknowns. The problem is readily solvable.
We will take the mass balance equation, rewrite the left-hand side so that the only
variable (other than hydroxide or hydronium) is [Pb2+] and the right-hand side so that the
only variable is [PO43-]. Then with the Ksp expression we can make the left-hand side
only a variable of [PO43-] also. Then we have 1 equation in 1 unknown (because we
know the hydroxide and hydronium and we just solve for each species by substitution.
2 6 Pb 2 + @ + K c 6Pb 2 + @ = 2 6 Pb 2 + @ 1 + K c
n 3
d
n
3d
6 H +@
6 H +@
= 6 PO 34 - @ +
K b1 6 PO3 - @ + K b1 K b2 6 PO 3 - @ + K b1 K b2 K b3 6 PO 3 - @
4
2
4
3
4
6OH - @
6OH - @
6OH - @
2 6 Pb 2 + @ d1 + K c n = 1 + K b1 + K b1 K b2 + K b1 K b2 K b3 6 PO 3 - @
d
n
3
4
3
6 H +@
6OH - @ 6OH - @ 2
6 OH - @
1
3
K sp
2
1 + K c+ n = d1 + K b1- + K b1 K- b22 + K b1 K b2- K3 b3 n 6 PO 34 - @
2
3 e 6 PO 34 - @ o d
6H @
6OH @ 6OH @
6OH @
This, upon rearrangement, will produce an expression which is fifth order in phosphate
and since we know the hydroxide and hydronium in concentrations (as well as the values
for the constants) we can solve this. In fact, this equation can be rather easily solved
analytically for the [PO43-] concentration. Gather the constants on one side, get rid of the
fractional power by cubing both sides, and gather the phosphate terms.
3 d1 + K b1- + K b1 K- b22 + K b1 K b2- K3 b3 n
1
3
6OH @ 6OH @
6OH @
K sp
6 PO34 - @
=
2o
e
3K
c
6 PO 4 @
2 d1 +
n
6 H +@
Z
_3
] 3 d1 + K b1 + K b1 K b22 + K b1 K b2 K3 b3 nb
]
b
6 OH - @ 6OH - @
6OH - @
3
K sp
` 6 PO 34 - @
2 = [
36 PO 4 @ ]
b
2 d1 + K c+ n
]
b
6H @
\
a
5
K
sp
6 PO 34 - @ = Z
_3
] 3 d1 + K b1 + K b1 K b22 + K b1 K b2 K3 b3 nb
]
b
6OH - @ 6OH - @
6OH - @
[
`
Kc
]
b
2
1
+
d
n
]
b
6 H +@
\
a
R
V 15
K sp
3S
6 PO 4 @ = Z
_ 3 WW
S]
K
K
K
b1
b1 K b2
b1 K b2 K b3 b
2 +
3
nb W
S ] 3 d1 + 6OH - @ + 6
6 OH - @
OH - @
S[
` W
S]
K
c
b W
2 d1 +
n
b W
S]
6 H +@
\
a X
T
While this may look imposing, we in fact know everything inside the brackets. It just a
little bit of arithmetic. If you use a spreadsheet to do the calculation, perhaps even doing
it in sections, you can go back and easily check the equation to make sure it is doing the
correct calculation.
When you do the calculation, you should find that [PO43-] = 1.11 x 10-14 M. With this and
the other two concentrations, we can easily solve for the other concentrations by
substitution.
All 8 species’s concentrations are:
[H+] = 3.16 x 10-6 M
[OH-] = 3.19 x 10-9 M
[PO43-] = 1.11 x 10-14 M
[Pb2+] = 6.25 x 10-6 M
[PbOH+] = 4.90 x 10-8 M
[HPO42-] = 8.27 x 10-8 M
[H2PO4-] = 4.12 x 10-6 M
[H3PO4] = 1.78 x 10-9 M