Calculate the pH of a Saturated Solution When Given the Ksp

Back to Equilibrium Menu

Return to a listing of many types of acid base problems and their solutions

Return to Acid Base menu


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

Warning: you need to know about Ksp 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 Ksp expression and then, at the end, we will use acid base concepts to get the pH.

Final Note: Ksp 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 Ksp that, for whatever reason, was not at 25.0 °C.


Problem #1: Calculate the pH of a saturated solution of AgOH, Ksp = 2.0 x 10¯8

Solution:

AgOH <===> Ag+ + OH¯

Ksp = [Ag+] [OH¯]

2.0 x 10¯8 = (x) (x)

x = 1.4 x 10¯4 M (this is the [OH¯])

pOH = - log [OH¯] = - log 1.4 x 10¯4 = 3.85

I actually took the negative log of 1.414 . . . x 10¯4; I did not use just 1.4

Because pH + pOH = 14, we have pH = 14 - 3.85 = 10.15


Problem #2: Calculate the pH of a saturated solution of Cu(OH)2, Ksp = 1.6 x 10¯19

Solution:

Cu(OH)2 <===> Cu2+ + 2 OH¯

Ksp = [Cu2+] [OH¯]2

1.6 x 10¯19 = (x) (2x)2 = 4x3

After dividing by 4 and then taking the cube root:

x = 3.42 x 10¯7 M

This is an important point: what we have calculated is 'x' and it is NOT the [OH¯]. That value is '2x.'

[OH¯] = 6.84 x 10¯7 M

pOH = - log 6.84 x 10¯7 = 6.165

pH = 14 - 6.165 = 7.835


Problem #3: Calculate the pH of a saturated solution of Mg(OH)2, Ksp = 5.61 x 10¯12

Solution:

Mg(OH)2 <===> Mg2+ + 2 OH¯

Ksp = [Mg2+] [OH¯]2

5.61 x 10¯12 = (x) (2x)2 = 4x3

After dividing by 4 and then taking the cube root:

x = 1.12 x 10¯4 M

This is an important point: what we have calculated is 'x' and it is NOT the [OH¯]. That value is '2x.'

[OH¯] = 2.24 x 10¯4 M

pOH = - log 2.24 x 10¯4 = 3.650

pH = 14 - 3.650 = 10.350


Problem #4: Calculate the pH of a saturated solution of Ba(OH)2, Ksp = 5.0 x 10¯3.

Solution:

5.0 x 10¯3 = (x) (2x)2

x = 0.10772 M

2x = [OH¯] = 0.21544 M

pOH = 10¯0.21544 = 0.67

pH = 14 - 0.67 = 13.33


The following three problems 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:

Problem #5: Ca(OH)2, Ksp = 7.9 x 10¯6 (pH = 12.10)

Problem #6: Mn(OH)2, Ksp = 4.6 x 10¯14 (no answer provided)

Problem #7: Ni(OH)2, Ksp = 2.8 x 10¯16 (no answer provided)


Back to Equilibrium Menu

Return to a listing of many types of acid base problems and their solutions

Return to Acid Base menu