Ksp: The Solubility-Product Constant
Source: http://www.chemteam.info/Equilibrium/Writing-Ksp-expression.html
This principle was first put forth by Walther Nernst in 1899. It has to do with solid substances usually considered insoluble in water. In each case, we will consider a saturated solution of the relatively insoluble substance that is in contact with some undissolved solid. Important points to consider are:
1) Some of the solid does dissolve. Not very much, but enough.
2) The substance dissociates upon dissolving.
3) There exists an equilibrium between the undissolved solid and the solvated ions. The solid is dissolving at the same rate that the solvated ions are precipitating back into solid form.
Since equilibrium principles can be used, that is where we start. Our first example is silver chloride, AgCl. When it dissolves, it dissociates like this:
AgCl (s) ⇋ Ag+ (aq) + Cl¯ (aq)
An equilibrium expression can be written:
Kc = ( [Ag+] [Cl¯] ) / [AgCl]
Now, we come to an important point. When the AgCl is enclosed in square brackets like this -- [AgCl] -- that means the "molar concentration" of solid AgCl. This value is a constant!! Why?
Answer: The "molar concentration" of a solid (it's not a useful chemistry idea, so it is seldom used) can be directly related to the density, which is also a constant. Here is a graphic which summarizes the relationship:
What we do is move the [AgCl] to the other side and incorporate it with the equilibrium constant.
Kc [AgCl] = [Ag+] [Cl¯]
Since Kc [AgCl] is a constant, we replace it with a single symbol Ksp with the “sp” standing for “solubility product”. Like this:
Ksp = [Ag+] [Cl¯]
(Just a side point - as you go on in chemistry, you'll get introduced to the concept of activity. The activity of a solid is defined as equal to the value of one. Since the activity of AgCl(s) = 1, it just drops out. However, like I said, activity is for the future. Not right now.)
It turns out that the Ksp value can be either directly measured or calculated from other experimental data. Knowing the Ksp, we can calculate the solubility of the substance in a very straightforward fashion.
Here are several other examples of dissociation equations and their Ksp expressions:
Sn(OH)2 (s) ⇋ Sn2+ (aq) + 2 OH¯ (aq) / Ksp = [Sn2+] [OH¯]2Ag2CrO4 (s) ⇋ 2 Ag+ (aq) + CrO42¯ (aq) / Ksp = [Ag+]2 [CrO42¯]
Fe(OH)3 (s) ⇋ Fe3+ (aq) + 3 OH¯ (aq) / Ksp = [Fe3+] [OH¯]3
In order to write Ksp expressions properly, you must know how ionic substances dissociate in water. That means, you have to know your chemical nomenclature, polyatomic ions, and the charges associated with ion.
Also, and this is important, so pardon the shouting:
EACH CONCENTRATION IN THE Ksp EXPRESSION IS RAISED TO THE POWER OF ITS COEFFICIENT IN THE BALANCED EQUATION.
Here are some practice problems for writing Ksp expressions. Write the chemical equation showing how the substance dissociates and write the Ksp expression (answers on next page):
1) AlPO4
2) BaSO4
3) CdS
4) Cu3(PO4)2
5) CuSCN
6) Hg2Br2
7) AgCN
8) Zn3(AsO4)2
9) Mn(IO3)2
10) PbBr2
11) SrCO3
12) Bi2S3
Answers:
Dissociation Equation Ksp expression
AlPO4 ⇋ Al3+ (aq) + PO43¯ (aq) Ksp = [Al3+] [PO43¯]
BaSO4 ⇋ Ba2+ (aq) + SO42¯ (aq) Ksp = [Ba2+] [SO42¯]
CdS ⇋ Cd2+ (aq) + S2¯ (aq) Ksp = [Cd2+] [S2¯]
Cu3(PO4)2 ⇋ 3 Cu2+ (aq) + 2 PO43¯ (aq) Ksp = [Cu2+]3 [PO43¯]2
CuSCN ⇋ Cu+ (aq) + SCN¯ (aq) Ksp = [Cu+] [SCN¯]
Hg2Br2 ⇋ Hg22+ (aq) + 2 Br¯ (aq) Ksp = [Hg22+] [Br¯]2
AgCN ⇋ Ag+ (aq) + CN¯ (aq) Ksp = [Ag+] [CN¯]
Zn3(AsO4)2 ⇋ 3 Zn2+ (aq) + 2 AsO43¯ (aq) Ksp = [Zn2+]3 [AsO43¯]2
Mn(IO3)2 ⇋ Mn2+ (aq) + 2 IO3¯ (aq) Ksp = [Mn2+] [IO3¯]2
PbBr2 ⇋ Pb2+ (aq) + 2 Br¯ (aq) Ksp = [Pb2+] [Br¯]2
SrCO3 ⇋ Sr2+ (aq) + CO32¯ (aq) Ksp = [Sr2+] [CO32¯]
Bi2S3 ⇋ 2 Bi3+ (aq) + 3 S2¯ (aq) Ksp = [Bi3+]2 [S2¯]3
Solving Ksp Problems: Part One
There are five substances I will use as examples. I will do AgCl in this section and do two others in each of part two and part three.
The problem will always be the same in each example. It is:
Calculate the molar solubility (in mol/L) of a saturated solution of the substance.
For example, Silver chloride, AgCl, has a Ksp = 1.77 x 10¯10. Calculate its solubility in moles per liter.
The dissociation equation is as before:
AgCl (s) ⇋ Ag+ (aq) + Cl¯ (aq)
The Ksp expression is as before:
Ksp = [Ag+] [Cl¯]
This is the equation we must solve. First we put in the Ksp value:
1.77 x 10¯10 = [Ag+] [Cl¯]
Now, we have to reason out the values of the two guys on the right, First of all, I have no idea what the values are so I'll use unknowns. Like this:
[Ag+] = x
[Cl¯] = y
However, I now have a BIG problem (No, not B.I.G., just BIG). The problem is that I have two unknowns, but only one equation. So it seems I'm stuck, untillllll . . .
I examine the chemical equation and see that there is a one-to-one ratio between Ag+ and Cl¯. I know this from the coefficients (both one) of the balanced equation. When AgCl dissociates it creates 1 Ag+ for every 1 Cl-.
That means that the concentrations of the two ions are EQUAL. I can use the same unknown to represent both. Like this:
[Ag+] = x = [Cl-]
Substituting, we get:
1.77 x 10¯10 = (x) (x) = x2
Now, we take the square root of both sides. I hope I'm not too insulting when I emphasize both sides. I have had lots of people take the square root of the x2 side, but not the other. After the square root, we get:
x = 1.33 x 10-5 M
This is the answer because there is a one-to-one relationship between the Ag+ dissolved and the AgCl it came from. So, the molar solubility of AgCl is 1.33 x 10¯5 moles per liter.
Do you want to see more problems like this?
OK, you asked for it.
The key point is that all these substances dissociate into a one-to-one ratio between the ions in solution. Here are the substances and their Ksp values:
AlPO4 / 9.83 x 10¯21BaSO4 / 1.07 x 10¯10
Once again, the standard problem is to calculate the molar solubility given the Ksp.
Aluminum phosphate
Here is the dissociation equation:
AlPO4 <===> Al3+ (aq) + PO43¯ (aq)
Here is the Ksp expression:
Ksp = [Al3+] [PO43¯]
Keep in mind that the key point is the one-to-one ratio of the ions in solution. This means that the two ions are equal in their concentration. That allows this equation:
9.83 x 10¯21 = x2
Solving gives:
x = 9.91 x 10¯11 M
which is the desired answer.
Barium sulfate
The dissociation equation and the Ksp expression:
BaSO4 <===> Ba2+ (aq) + SO42¯ (aq)
Ksp = [Ba2+] [SO42¯]
The equation:
1.07 x 10¯10 = x2
when solved, gives:
x = 1.03 x 10¯5 M
which is the desired answer.
Remember, this is the desired answer because the dissolved ions and the solid are also in a one-to-one molar ratio.
Solving Ksp Problems: Part Two
As in each part, the problem is always the same in each example. It is: Calculate the molar solubility (in mol/L) of a saturated solution of the substance. However, there is additional explaining to do when compared to the AgCl example.
Here are the two substances:
Sn(OH)2 / Ksp = 5.45 x 10¯27Ag2CrO4 / Ksp = 1.12 x 10¯12
Tin(II) hydroxide
Here is the equation for dissociation:
Sn(OH)2 ⇋ Sn2+ + 2 OH¯
and here is the Ksp expression:
Ksp = [Sn2+] [OH¯]2
So far, nothing out of the ordinary. However, that two in front of the hydroxide will come into play real soon.
As in right now!!
The ratio between Sn2+ and OH¯ is one-to-two. That means that however much Sn2+ dissolves, we get DOUBLE that amount of OH¯. This is important, so go slow and think it through.
One Sn2+ makes two OH¯. That means that if 'x' Sn2+ dissolves, then '2x' of the OH¯ had to have dissolved.
Boy, I hope you got that!! Let's move on.
We now write an equation:
5.45 x 10¯27 = (x) (2x)2
Wait, Mr. Chemistry person, why did you DOUBLE the concentration of hydroxide?
I didn't. The concentration of OH¯ is equal to double the concentration of Sn2+. Let's move on.
Oops! Important point. Don't write 2x2. That's wrong! It's (2x)2; two x the quantity squared.
So we have:
4x3 = 5.45 x 10¯27
and solving that, we get:
x = 1.11 x 10¯9 M
Now, this is the answer because there is a one-to-one ratio between Sn2+ and Sn(OH)2
Silver chromate
Here is the usual info:
Ag2CrO4 ⇋ 2 Ag+ + CrO42¯
Ksp = [Ag+]2 [CrO42¯] = 1.12 x 10¯12
In this problem the cofficient of 2 is on the first ion, not the second. No problem!
1.12 x 10¯12 = (2x)2 (x)
In other words, what I'm doing is allowing x to equal the concentration of whatever ion has a one in front of it. That ion will always be in a one-to-one ratio with the solid which is dissolving. So solving for x gives me the molar solubility of the substance.
We now have:
4x3 = 1.12 x 10¯12
which is solved to give the answer:
x = 6.54 x 10¯5 M
Do you want to see more problems like these?
Solving Ksp Problems: Part Three
As in the other parts the problem stays the same in each example. It is: Calculate the molar solubility (in mol/L) of a saturated solution of the substance.
I'm going to assume you've gone through the other files first, so the presentations here will be a bit abbreviated. I'm hoping that you have begun to see a pattern developing. If not, the pattern is that the coefficient winds up doing two things:
1) it puts a power on the 'x' which represents that particular ion and
2) it puts a coefficient in front of the 'x'
The two example substances are:
Fe(OH)3 / Ksp = 2.64 x 10¯39Ag3PO4 / Ksp = 8.88 x 10¯17
Iron(III) hydroxide
Fe(OH)3 ⇋ Fe3+ + 3 OH¯
Ksp = [Fe3+] [OH¯]3
2.64 x 10¯39 = (x) (3x)3
27x4 = 2.64 x 10¯39
x = 9.94 x 10¯11 M
How's that for abbreviated? The 3x comes from the fact that there are three hydroxides produced for every Fe3+ ion, but you knew that, didn't you?
Silver phosphate
Ag3PO4 ⇋ 3 Ag+ + PO43¯
Ksp = [Ag+]3 [PO43¯]
8.88 x 10¯17 = (3x)3 (x)
27x4 = 8.88 x 10¯17
x = 4.26 x 10¯5 M
Solving Ksp Problems: Part Four
As in the other parts the problem stays the same in each example. It is: Calculate the molar solubility (in mol/L) of a saturated solution of the substance.
The two example substances are:
Bi2S3 / Ksp = 1.82 x 10¯99Cu3(PO4)2 / Ksp = 1.93 x 10¯37
Bismuth sulfide
Bi2S3 ⇋ 2 Bi3+ + 3 S2¯
Ksp = [Bi3+]2 [S2¯]3
1.82 x 10¯99 = (2x)2 (3x)3
108x5 = 1.82 x 10¯99
x = 7.00 x 10¯21 M
Note how 'x' still is the moles of the substance that dissolved.
There is also a bit of a problem with this problem. When you divide the Ksp by 108, the answer is 1.685 x 10¯101. Some calculators are not equipped to handle three digit exponents. However, bring up a spreadsheet on your computer and you will be able to enter values and a formula to calculate the value.
Copper(II) phosphate
Cu3(PO4)2 ⇋ 3 Cu2+ + 2 PO43¯
Ksp = [Cu2+]3 [PO43¯]2
1.93 x 10¯37 = (3x)3 (2x)2
108x5 = 1.93 x 10¯37
x = 1.78 x 10¯8 M
Calculating the Ksp from the Molar Solubility
The molar solubility of a substance is the number of moles that dissolve per liter of solution. For very soluble substances (like sodium nitrate, NaNO3), this value can be quite high, exceeding 10.0 moles per liter of solution in some cases.
For insoluble substances like silver bromide (AgBr), the molar solubility can be quite small. In the case of AgBr, the value is 5.71 x 10¯7 moles per liter.
Given this value, how dos one go about calculating the Ksp of the substance? Here is a skeleton outline of the process:
1) Write the chemical equation for the substance dissolving and dissociating.
2) Write the Ksp expression.
3) Insert the concentration of each ion and multiply out.
Problem #1: Determine the Ksp of silver bromide, given that its molar solubility is 5.71 x 10¯7 moles per liter.
When AgBr dissolves, it dissociates like this:
AgBr (s) ⇋ Ag+ (aq) + Br¯ (aq)
The Ksp expression is:
Ksp = [Ag+] [Br¯]
There is a 1:1 ratio between AgBr and Ag+ and there is a 1:1 ratio between AgBr and Br¯. This means that, when 5.71 x 10¯7 mole per liter of AgBr dissolves, it produces 5.71 x 10¯7 moles per liter of Ag+ and 5.71 x 10¯7 moles per liter of Br¯ in solution.
Putting the values into the Ksp expression, we obtain:
Ksp = (5.71 x 10¯7) (5.71 x 10¯7) = 3.26 x 10¯13
Video: Calculating the Ksp from the Molar Solubility
Problem #2: Determine the Ksp of calcium fluoride (CaF2), given that its molar solubility is 2.14 x 10¯4 moles per liter.
When CaF2 dissolves, it dissociates like this:
CaF2 (s) ⇋ Ca2+ (aq) + 2 F¯ (aq)
The Ksp expression is:
Ksp = [Ca2+] [F¯]2
There is a 1:1 ratio between CaF2 and Ca2+, BUT there is a 1:2 ratio between CaF2 and F¯. This means that, when 2.14 x 10¯4 mole per liter of CaF2 dissolves, it produces 2.14 x 10¯4 moles per liter of Ca2+, BUT 4.28 x 10¯4 moles per liter of F¯ in solution.
Putting the values into the Ksp expression, we obtain:
Ksp = (2.14 x 10¯4) (4.28 x 10¯4)2 = 3.92 x 10¯11
Please note, I DID NOT double the F¯ concentration. I took the Ca2+ concentration and doubled it to get the F¯ concentration.