Solving Ksp Problems: Part One - s2

Solving Ksp Problems:
Part Two - 4s3
Solving Ksp Problems:
Part Three - 27s4
Solving Ksp Problems:
Part Four - 108s5
Solving Ksp Problems:
Part Five - 256s5
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The general problem is this:

Given the Ksp, calculate the molar solubility (in mol/L) of a saturated solution of the substance.

Example #1: Silver chloride, AgCl, has a Ksp = 1.77 x 10¯10. Calculate its solubility in moles per liter.


1) Write the dissociation equation:

AgCl(s) ⇌ Ag+(aq) + Cl¯(aq)

2) Write the Ksp expression:

Ksp = [Ag+] [Cl¯]

This is the equation we must solve.

3) Put in the Ksp value:

1.77 x 10¯10 = [Ag+] [Cl¯]

4) We have to reason out the values of the two guys on the right. I do that by first assigning a variable, s, to the molar solubility of AgCl. That's the value that I want to determine.

To do this, I look at the molar relationship between AgCl and Ag+. (I do this because I want to express the concentrations of the ions by using 's.') I note that the molar relationship is a 1:1 molar ratio, meaning that for every one mole of AgCl that dissolves, one mole of Ag+ is produced. That leads me to this:
[Ag+] = s

Now, I look at the relationship between AgCl and Cl¯. I see that it is also a 1:1 molar ratio, leading me to this:

[Cl¯] = s

I am now ready to substitute into the Ksp expression.

5) Substituting, we get:

1.77 x 10¯10 = (s) (s)

7) 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 in my classes take the square root of the x2 side, but not the other. After the square root, we get:

s = 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.

One last thing. The Ksp value does not have any units on it, but when you get to the value for s, be sure to put M (for molarity) on it. The reasons behind this are complex and beyond the scope of the ChemTeam's goals for this web site.

Comment: it's important that you know how substances ionize. AgCl is easy, but you may not know that CuSCN (example #4) and silver azide (example #5) also ionize in a 1:1 molar ratio like AgCl.

Warning: You may know lots of common ions but, in a problem like the one under discussion, you may get an unusual one thrown at you on the test. Be prepared!

Example #2: Aluminum phosphate has a Ksp of 9.83 x 10¯21. What is its molar solubility in pure water?


1) Here is the dissociation equation:

AlPO4(s) ⇌ Al3+(aq) + PO43¯(aq)

2) Here is the Ksp expression:

Ksp = [Al3+] [PO43¯]

3) 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 = (s) (s)

4) Solving gives:

s = 9.91 x 10¯11 M

which is the answer.

Example #3: Calculate the molar solubility of barium sulfate, Ksp = 1.07 x 10¯10


1) The dissociation equation and the Ksp expression:

BaSO4(s) ⇌ Ba2+(aq) + SO42¯(aq)
Ksp = [Ba2+] [SO42¯]

2) The equation:

1.07 x 10¯10 = (s) (s)

3) when solved, gives:

s = 1.03 x 10¯5 M

The answer.

Remember, this is the answer because the dissolved ions and the solid are also in a one-to-one molar ratio.

Notice how I did not say 'saturated solution' in the problem. When you see this, you need to assume that it is a saturated solution. Anything else makes the problem unworkable and that is not the intent of the question writer.

Example #4: CuSCN, Ksp = 1.77 x 10¯13

CuSCN(s) ⇌ Cu+(aq) + SCN¯(aq)

molar solubility = 4.21 x 10¯7 M

Example #5: Silver azide has the formula AgN3 and Ksp = 2.0 x 10¯8

AgN3(s) ⇌ Ag+(aq) + N3¯(aq)

Note azide, a fairly uncommon polyatomic ion. It could show up on the test! Calculating the molar solubility is left to the student.

Example #6: Calculate the [Ba2+] if the [SO42¯] = 0.0153 M. Ksp = 1.07 x 10¯10


1) The compound in solution is BaSO4. This dissociates:

BaSO4(s) ⇌ Ba2+(aq) + SO42¯(aq)

2) The Ksp expression is:

Ksp = [Ba2+] [SO42¯]

3) Substitute and solve:

1.07 x 10¯10 = (s) (0.0153 + s)

s = 1.07 x 10¯10 / 0.0153

s = 6.99 x 10¯9 M

Note how I ignored the s in (0.0153 + s). That's because s is very small compared to 0.0153. By the way, the sulfate already present in solution came from some other sulfate-containing compound, say sodium sulfate, Na2SO4. We will completely ignore the sodium ion, because it plays no role in the Ksp equilibrium.

Bonus Example: Magnesium ammonium phosphate, MgNH4PO4, Ksp = 2.5 x 10¯13

This is how the substance ionizes:

MgNH4PO4(s) ⇌ Mg2+(aq) + NH4PO42¯(aq)

The ammonium phosphate polyatomic ion, NH4PO42¯, is an uncommon one.

Solving Ksp Problems:
Part Two - 4s3
Solving Ksp Problems:
Part Three - 27s4
Solving Ksp Problems:
Part Four - 108s5
Solving Ksp Problems:
Part Five - 256s5
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