The general problem is this:

Given the K_{sp}, calculate the molar solubility (in mol/L) of a saturated solution of the substance.

**Example #1:** Silver chloride, AgCl, has a K_{sp} = 1.77 x 10¯^{10}. Calculate its solubility in moles per liter.

**Solution:**

1) Write the dissociation equation:

AgCl(s) ⇌ Ag^{+}(aq) + Cl¯(aq)

2) Write the K_{sp} expression:

K_{sp}= [Ag^{+}] [Cl¯]This is the equation we must solve.

3) Put in the K_{sp} 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^{+}] = sNow, 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¯] = sI am now ready to substitute into the K

_{sp}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 s^{2} 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 K_{sp} 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. Also, there are teachers that insist on units for the K_{sp}. If you have one of them, please follow their lead. Do not tell them that some guy (me!) on the Internet says they are wrong. Thanks.

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 K_{sp} of 9.83 x 10¯^{21}. What is its molar solubility in pure water?

**Solution:**

1) Here is the dissociation equation:

AlPO_{4}(s) ⇌ Al^{3+}(aq) + PO_{4}^{3}¯(aq)

2) Here is the K_{sp} expression:

K_{sp}= [Al^{3+}] [PO_{4}^{3}¯]

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

**Example #3:** Calculate the molar solubility of barium sulfate, K_{sp} = 1.07 x 10¯^{10}

**Solution:**

1) The dissociation equation and the K_{sp} expression:

BaSO_{4}(s) ⇌ Ba^{2+}(aq) + SO_{4}^{2}¯(aq)K

_{sp}= [Ba^{2+}] [SO_{4}^{2}¯]

2) The equation:

1.07 x 10¯^{10}= (s) (s)

3) When solved, gives:

s = 1.03 x 10¯^{5}M

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:** Calculate the molar solubility of CuSCN, K_{sp} = 1.77 x 10¯^{13}

CuSCN(s) ⇌ Cu^{+}(aq) + SCN¯(aq)K

_{sp}= [Cu^{+}] [SCN¯]1.77 x 10¯

^{13}= (s) (s)Molar solubility = 4.21 x 10¯

^{7}M

**Example #5:** Silver azide has the formula AgN_{3} and K_{sp} = 2.0 x 10¯^{8}. Determine the molar solubility of silver azide.

AgN_{3}(s) ⇌ Ag^{+}(aq) + N_{3}¯(aq)K

_{sp}= [Ag^{+}] [N_{3}¯]2.0 x 10¯

^{8}= (s) (s)Azide, a fairly uncommon polyatomic ion, is a good example of what might show up on the test. Calculating the molar solubility is left to the student.

**Example #6:** Calculate the molar solubility of CdS, K_{sp} = 1.40 x 10¯^{29}

**Solution:**

CdS(s) ⇌ Cd^{2+}(aq) + S^{2}¯(aq)K

_{sp}= [Cd^{2+}] [S^{2}¯]1.40 x 10¯

^{29}= (s) (s)s = 3.74 x 10¯

^{15}M

**Example #7:** Calculate the molar solubility of mercury(I) sulfate, Hg_{2}SO_{4}, K_{sp} = 6.50 x 10¯^{7}

**Solution:**

Hg_{2}SO_{4}(s) ⇌ Hg_{2}^{2+}(aq) + SO_{4}^{2}¯(aq)K

_{sp}= [Hg_{2}^{2+}] [SO_{4}^{2}¯}6.50 x 10¯

^{7}= (s) (s)s = 0.000806 M

Writing the mercury(I) ion as Hg

^{+}is incorrect. The proper way is as above, Hg_{2}^{2+}.

**Bonus Example:** Magnesium ammonium phosphate, MgNH_{4}PO_{4}, K_{sp} = 2.5 x 10¯^{13}

This is how the substance ionizes:

MgNH_{4}PO_{4}(s) ⇌ Mg^{2+}(aq) + NH_{4}PO_{4}^{2}¯(aq)

The ammonium phosphate polyatomic ion, NH_{4}PO_{4}^{2}¯, is an uncommon one.

The molar solubility magnesium ammonium phosphate is 5.0 x 10¯^{6} M. Verification of this is left to the student.