### What is the minimum pH required for precipitation?

Problem #1: What is the minimum pH at which Cr(OH)3 will precipitate? Ksp of Cr(OH)3 is 6.70 x 10¯31

Solution:

1) Write the dissociation equation:

Cr(OH)3 ⇌ Cr3+ + 3OH¯

2) Write the Ksp expression:

Ksp = [Cr3+] [OH¯]3

3) Plug into the Ksp expression:

6.70 x 10¯31 = (s) (3s)3

4) Solve for [OH¯]:

[Cr3+] = s = 1.255 x 10¯8 M (I kept some guard digits)
[OH¯] = 3s = 3.765 x 10¯8 M

5) Calculate the pH given the [OH¯]:

pOH = -log 3.765 x 10¯8 = 7.424
pH = 14 - pOH = 14 - 7.424 = 6.576

Note that even in a slightly acidic environment, Cr(OH)3 will start to precipitate.

Problem #2: What is the minimum pH at which Cr(OH)3 will precipitate if the solution has [Cr3+] = 0.0670 M? Ksp of Cr(OH)3 is 6.70 x 10¯31

Solution:

1) Write the dissociation equation:

Cr(OH)3 ⇌ Cr3+ + 3OH¯

2) Write the Ksp expression:

Ksp = [Cr3+] [OH¯]3

3) Plug into the Ksp expression:

6.70 x 10¯31 = (0.0670) (s)3

Note that 3s is not necessary. In example one, s was assigned to the chromium ion and we knew the hydroxide to be three times greater than that value. Here, the hydroxide is simply an unknown value and it is NOT expressed in terms of some other unknown value (as it was in example #1).

4) Solve for s, which is the [OH¯]:

s = 2.1544 x 10¯10 M

5) Calculate the pH:

pH = 14 - pOH = 14 - 9.667 = 4.333

Problem #3: At what pH will Al(OH)3(s) begin to precipitate from 0.10 M AlCl3? The Ksp of Al(OH)3 is 1.90 x 10¯33

Solution:

1) Write the dissociation equation:

Al(OH)3 ⇌ Al3+ + 3OH¯

2) Write the Ksp expression:

Ksp = [Al3+] [OH¯]3

3) Plug into the Ksp expression:

1.90 x 10¯33 = (0.10) (s)3

4) Solve for s, which is the [OH¯]:

s = 2.6684 x 10¯11 M

5) Calculate the pH:

pH = 14 - pOH = 14 - 10.574 = 3.426

In other words, AlCl3 will be soluble only in fairly acidic solutions. In this particular example, Al(OH)3 will precipitate if the pH is 3.426 or higher. The pH MUST be maintained at 3.426 or lower in order to keep the AlCl3 in solution.