Please be aware that this problem does not require a concentration to be given in the problem. We know the K_{sp} and we know the solution is saturated. This information will be sufficient.

Warning: you need to know about K_{sp} AND acid base ideas to do this problem type. If you lack one or the other of these skills, just be aware you just might struggle a little in your understanding of this problem type.

To solve the problem, we must first calculate the [OH¯]. To do this, we will use the K_{sp} expression and then, at the end, we will use acid base concepts to get the pH.

Final Note: K_{sp} are almost always given at 25.0 °C in reference sources. All problems in this tutorial are taken to be at 25.0 °C. If you were to see a problem where the specified temperature was different, it's probably just that the reference source gave a K_{sp} that, for whatever reason, was not at 25.0 °C.

**Example #1:** Calculate the pH of a saturated solution of AgOH, K_{sp} = 2.0 x 10¯^{8}

**Solution:**

AgOH ⇌ Ag^{+}+ OH¯K

_{sp}= [Ag^{+}] [OH¯]2.0 x 10¯

^{8}= (s) (s)s = 1.4 x 10¯

^{4}M (this is the [OH¯])pOH = - log [OH¯] = - log 1.4 x 10¯

^{4}= 3.85I actually took the negative log of 1.414 . . . x 10¯

^{4}; I did not use just 1.4Because pH + pOH = 14, we have pH = 14 - 3.85 = 10.15

**Example #2:** Calculate the pH of a saturated solution of Cu(OH)_{2}, K_{sp} = 1.6 x 10¯^{19}

**Solution:**

Cu(OH)_{2}⇌ Cu^{2+}+ 2 OH¯K

_{sp}= [Cu^{2+}] [OH¯]^{2}1.6 x 10¯

^{19}= (s) (2s)^{2}= 4s^{3}After dividing by 4 and then taking the cube root:

s = 3.42 x 10¯

^{7}MThis is an important point: what we have calculated is 's' and it is NOT the [OH¯]. That value is '2s.'

[OH¯] = 6.84 x 10¯

^{7}MpOH = - log 6.84 x 10¯

^{7}= 6.165pH = 14 - 6.165 = 7.835

**Example #3:** Calculate the pH of a saturated solution of Mg(OH)_{2}, K_{sp} = 5.61 x 10¯^{12}

**Solution:**

Mg(OH)_{2}⇌ Mg^{2+}+ 2 OH¯K

_{sp}= [Mg^{2+}] [OH¯]^{2}5.61 x 10¯

^{12}= (s) (2s)^{2}= 4s^{3}After dividing by 4 and then taking the cube root:

s = 1.12 x 10¯

^{4}MThis is an important point: what we have calculated is 's' and it is NOT the [OH¯]. That value is '2s.'

[OH¯] = 2.24 x 10¯

^{4}MpOH = - log 2.24 x 10¯

^{4}= 3.650pH = 14 - 3.650 = 10.350

**Example #4:** Calculate the pH of a saturated solution of Ba(OH)_{2}, K_{sp} = 5.0 x 10¯^{3}.

**Solution:**

5.0 x 10¯^{3}= (s) (2s)^{2}s = 0.10772 M

2s = [OH¯] = 0.21544 M

pOH = 10¯

^{0.21544}= 0.67pH = 14 - 0.67 = 13.33

The following three examples are all of the form X(OH)_{2}. These are the ones most commonly asked on tests and in worksheets. Calculate the pH of a saturated solution of:

**Example #5:** Ca(OH)_{2}, K_{sp} = 7.9 x 10¯^{6} (pH = 12.10)

**Example #6:** Mn(OH)_{2}, K_{sp} = 4.6 x 10¯^{14} (no answer provided)

**Example #7:** Ni(OH)_{2}, K_{sp} = 2.8 x 10¯^{16} (no answer provided)