In the last unit we reviewed the common types of equilibrium expressions.
The examples typically dealt with relatively simple systems with only a single equilibrium to work with.
It is much more common to have two or more simultaneous equilibria.
This unit reviews the approaches taken to solve more complex problems.
The best way to learn how to work these problems is to do a series of examples.
We’ll attempt to follow each step while covering a range of problems.
Solubility, acid-base and complex formation will all be involved at one point or another.
Determine the concentration of silver ion when excess AgCl is added to 0.1M NaBr.
We are adding a relative insoluble material to a solution that contains an anion that also produces a precipitate with silver.
We know that [Na+] = 0.1 M We want to know [Ag+]
Lets write down the equilibria involved.
AgCl(s) Ag+ + Cl- AgBr(s) Ag+ + Br-
KSPAgBr = [Ag+][Br-] = 4.9x10-13 KSPAgCl = [Ag+][Cl-] = 1.8x10-10
Its reasonable to assume that as silver enters the system, some will precipitate as AgBr.
OK, we have 4 species: Na+, Ag+, Cl- and Br- We know that Na+ = 0.1M since there is no
reason for it to change.
We have three unknowns but only two equations so far. We need to consider mass or charge balances.
Lets see what we know about the system.
First, every silver that enters the solution brings a chloride with it.
However, some silver is expected to precipitate as AgBr.
The total charge of the system must equal 0.
so [ Na+ ] + [ Ag+] = [ Cl- ] + [ Br- ]
We know that [Na+] = 0.1 so 0.1 + [Ag+] = [Cl-] + [Br-]
So, we now have three equations and three unknowns -- we’re ready to solve this thing.
We now need to express a single equation in terms of one unknown.
We’ll use our mass balance
0.1 + [Ag+] = [Cl-] + [Br-]
Br- and Cl- can be expressed in terms of silver using our KSP expressions:
[Br-] = KSPAgBr / [Ag+] [Cl-] = KSPAgCl / [Ag+]
0.1 + [Ag+]= KSPAgBr / [Ag+] + KSPAgCl / [Ag+]
We can rearrange this to:
[Ag+]2 + 0.1 [Ag+] - KSPAgBr -KSPAgCl This quadratic expression can be written as
X2 + 0.1 X - 1.805 x10-10
Solve using the quadratic expression.
Solving for X gives us [Ag+] = 1.081 x 10-9 M
No approximations were required for this solution.
-b + b2 - 4ac X = 2a
Lets make things a bit more challenging.
Calculate the equilibrium concentration of silver ion in 1.000 M NH3 if the solution also contains 0.1 moles of silver nitrate.
You may assume that no conversion of ammonia to ammonium occurs.
This problem stepwise formation constants.
Lets start by setting up the equilibria and seeing what we know about the system.
Ag+ + NH3 Ag(NH3)+
+ NH3 Ag(NH3)2+
K1
K2
We can see that there are two equilibrium expressions.
Now on to writing out each expression and looking up the various K values.
K1 = 2.34x103 =
K2 = 6.90x103 =
[Ag(NH3)+] [Ag+] [NH3]
[Ag(NH3)2+] [Ag(NH3)+] [NH3]
We now have two equations but 4 unknowns - 3 silver species and ammonia.
Time to look a mass/charge balances.
We know that the total silver concentration must be 0.1 M so:
0.1 = [Ag+] + [Ag(NH3)+] + [Ag(NH3)2+] We can also establish a second mass balance
based on ammonia:
[NH3] = 1.0 - [Ag(NH3)+] - 2 [Ag(NH3)2+] since any complex that forms will reduce the
ammonia concentration.
Next, can we assume that any species would be insignificant in our mass balances.
Here, we can look at our formation constants.
K1 = 2.34x103 = K2 = 6.90x103 =
[Ag(NH3)+] [Ag+] [NH3]
[Ag(NH3)2+] [Ag(NH3)+] [NH3] This shows that there are 2340 [Ag(NH3)+] for [Ag+] and 6900 [Ag(NH3)2+] for each[Ag (NH3)+].
Since almost all of our silver exists as [Ag(NH3)
2+], our mass balances become:
0.1 = [Ag(NH3)2+] [NH3] = 1.0 - 2 [Ag(NH3)2+]
= 0.8
So we know know the the values for two of our species. The other silver species are relatively simple to solve for.
[Ag(NH3)+] = [Ag(NH3)2+] 6.90x103 [NH3]
0.1 6.90x103 x 0.8
= 1.84x10-5
=
[Ag+] = [Ag(NH3)+] 2.34x103 [NH3]
1.84x10-5 2.34x103 x 0.8
= 9.82 x10-9
=
Lets take the last problem and bump it up a level in difficulty.
Determine the solubility of AgCl in a 1 M ammonia solution.
Again, you can assume that there is no significant conversion of ammonia to ammonium.
Lets look at our system.
AgCl(s) Ag++ Cl- + NH3 Ag(NH3)+
+ NH3 Ag(NH3)2+
K1
K2
We can see that there are now three equilibrium
expressions.
Now on to writing out each expression and looking up the various K values.
KSP
K1 = 2.34x103 = [Ag(NH3)+] [Ag+] [NH3] K2 = 6.90x103 = [Ag(NH3)2+]
[Ag(NH3)+] [NH3] KSP = 1.0 x 10-10 = [Ag+][Cl-]
We know from the last example that most of our silver will exist as the diamino complex.
We can derive a new expression by combining K1 and K2 which will be easier to work with.
K1K2= 1.6 x107 = [Ag(NH3)2+] [Ag+] [NH3]2
We now need mass or charge balances.
One can be developed for chloride:
[Cl-] = [Ag+] + [Ag(NH3)2+] And another for ammonia:
[NH3] = 1.00 - 2 [Ag(NH3)2+]
Almost all of our silver exists as [Ag(NH3)2+] so:
[NH3] = 1.00 - 2 [Cl-]
Solve for Cl-, it directly reflects the solubility of AgCl.
Need to get one equation in terms of [Cl-].
Let’s use our chloride mass balance:
[Cl-] = [Ag+] + [Ag(NH3)2+] First, [Ag+] = KSP / [Cl-] so:
[Cl-] = KSP / [Cl-] + [Ag(NH3)2+] Next, we can substitute our combined
equilibrium expression for [Ag(NH3)2+].
[Ag(NH3)2+] =K1K2[Ag+] [NH3]2 [Cl-] = KSP / [Cl-] + K1K2[Ag+] [NH3]2
[Cl-] = KSP / [Cl-] + K1K2KSP [NH3]2 / [Cl-] Again, we can replace [Ag+] with KSP / [Cl-].
Finally, we can use our other mass balance to replace ammonia with chloride based expression:
[Cl-] = KSP / [Cl-] + K1K2KSP [NH3]2 / [Cl-] [NH3] = 1.00 - 2 [Cl-]
[Cl-] = KSP / [Cl-] + K1K2KSP(1.00-2[Cl-])2 / [Cl-] Now clean this expression up.
[Cl-]2 = KSP+KSPK1K2(1-2[Cl-])2
[Cl-] =
= 0.121 M KSP+KSPK1K2
1 + 2KSPK1K2
Now for a similar problem that involves weak acids.
Determine the solubility of calcium oxalate at pH 4.00
As with our other examples, we need to start by writing down the entire equilibrium.
CaC2O4 Ca2+ + C2O42- + H3O+ HC2O4-
+ H3O+ H2C2O4 KSP
KA1 KA2
Here you can see that we must determine the [Ca+] to be able to calculate the solubility.
At least we know the pH.
We’ll use some shorthand. It will make things easier to type.
[H3O+] = [H]
[Ca2+] = [Ca]
[C2O42-] = [Ox]
[HC2O4-] = [HOx]
[H2C2O4] = [H2Ox]
Now to write out our equilibrium expressions.
KSP = [Ca][OX] = 1.6 x 10-8
KA1 = = 8.8 x 10-2
KA2 = = 5.1 x10-5
[H] = 1.00 x10-4
On to looking for any mass or charge balances.
[H][HOx]
[H2Ox]
[H][Ox]
[HOx]
We know that the calcium concentration should equal the total of all oxalate species so:
[Ca] = [H2Ox] + [HOx] + [Ox]
We now can solve the problem but should attempt to simplify the mass balance to make our life easier.
Let’s look at the oxalate species.
KA1 [H]
KA2 [H]
[HOX]
[H2Ox]
[Ox]
[HOx]
= 880
= 0.51
=
=
At pH 4.00, we have almost no H
2Ox.
Both HOx and Ox are present at a ratio of about 2:1 respectively.
Our mass balance becomes:
[Ca] = [HOx] + [Ox]
We now should attempt to solve for calcium.
All we need to do is use a series of substitutions to express our mass balance in terms of a single variable
- [Ca] is the best choice since it is a direct measure of our solubility.
[Ca] = [HOx] + [Ox]
[Ox] = KSP / [Ca]
[HOx] = [Ox][H] / KA2 = 1.96 [Ox]
[Ca] = 1.96 [Ox] + [Ox]
= 2.96 [Ox]
= 2.96 x KSP / [Ca]
OK, we now can solve for [Ca].
[Ca]2 = 2.96 x KSP [Ca] = 2.96 x KSP
= 2.18 x 10-4 M Not too bad!
Now you’re ready for a really ‘fun’ problem.
Calculate the KSP of MgNH4PO4 if you determine the solubility to be 0.52 g/l at pH 10.2
This looks pretty easy.
The molecular weight is 137.32 g/mol so you have 0.00379 moles.
The KSP expression is KSP = [Mg2+][NH4+][PO43-]
The H3O+ have been omitted from phosphate equilibria for clarity.
There’s obviously more to this than we first thought!
MgNH4PO4
Mg2+ + NH4+ + PO43- +
OH- Mg(OH)2
+ OH- NH3 + H2O
HPO42- H2PO4- H3PO4
KSP
KSP
KB
KA3 KA2 KA1
As bad as things look, we still can take the same approach to solve this littel jewel.
To determine the KSP, we’ll need to calculate the actual concentrations using other equilibrium expressions.
These are all [H3O+] or [OH-] based.
Fortunately, we know the pH.
KSPMg(OH)2 =1.2x10-11 = [Mg2+][OH-]2 KBNH4+ = 1.78x10-5 =
For phosphate KA1 = 7.5x10-3 = KA2 = 6.0x10-8 = KA3 = 4.8x10-13 =
[OH-][NH4+] [NH3]
[H3O+][H2PO4-] [H3PO4] [H3O+][HPO42-]
[H2PO4-] [H3O+][PO43-]
[HPO42-]
What do we already know?
[H3O+] = 3.98x10-11 [OH-] = 2.51x10-4
We also have several mass balances 3.79x10-3 = [NH4+] + [NH3]
= [H3PO4]+[H2PO4-]+[HPO42-]+[PO43-] This actually works out to be three problems in
one. We’ll start with Mg2+.
For Mg2+ we can see if we’ve exceeded the KSP for Mg(OH)2.
KSP =1.2 x 10-11 = [Mg2+][OH-]2 [Mg2+] = KSP/[OH-]2 = 4.7x10-4
We put 3.79x10-3 into the solution so some must have precipitated. The [Mg2+] is then based on the KSP for Mg(OH)2.
One down, two to go!
For ammonium, nothing will precipitate out of solution. However, some of it can be expected to be converted to ammonia.
[OH-][NH4+] [NH3] KBNH4+ = 1.78x10-5 =
KB
[OH-] = [NH4+]
[NH3] = 0.11
[NH4+]
0.11 = [NH3] Our mass balance for ammonia species is 3.79x10-3 = [NH4+] + [NH3]
Substituting for ammonia using:
Gives us [NH4+] + [NH4+]
0.11 = 3.79x10-3 [NH4+] = 3.8x10-4 M
Two down, one to go!
For phosphate, we need to determine which species exist at significant levels at pH 10.2
[H2PO4-] [H3PO4] KA1
[H3O+] KA2 [H3O+]
KA3 [H3O+]
[HPO42-] [H2PO4-] [PO43-] [HPO42-]
=
=
=
= 0.012
= 1.51x103
= 1.88x108
It appears that most of our phosphate exists as HPO42-
We can now solve for phosphate.
[HPO42-] = 3.79 x 10-3 M [PO43-] = 0.012 [HPO42-]
= 4.6x 10-5M KSP MgNH4PO4
= [Mg2+][NH4][PO4]
= (4.7x10-4)(3.8x10-4)(4.6x10-5) = 8.2x10-12
