Question

If aluminum hydroxide is added to an acidic solution with a pH of 1.93, what is the new pH after equilibrium is established? This problem is a bit tricky. So, extra help is given. The following ICE Table is useful: ...

Al(OH)3(s) ↔ Al3+(aq) + 3OH-(aq) Ksp = 2.0 x10-32 M4

I: ...Excess ... ...0 ......... ...from pH

C:.....-s....... ......s ...... ........3s

E:...Excess ... ...s.......... .... ? (can't add)

Hint, hydroxide produced neutralizes acid. [H+] =[H+]o -3s for 3s < [H+]o. From [H+] hydroxide concentration may be obtained. Math is simple, but you need to notice what reasonable approximation is useful.

Answer 1

Ksp of Al(OH)3 = 2.0 x 10^-32

Ksp = [Al3+][OH-]^3

pH = 1.93

pOH = 14 - pH = 12.07

pOH = -log[OH-]

[OH-] = 8.5 x 10^-13 M

                 Al(OH)3(s) <===> Al3+(aq) + 3OH-(aq)

I                      ex                      -              8.5 x 10^-13                     

C                    -x                      +x                +3x

E                     ex                       x            8.5 x 10^-13 + 3x

So,

2 x 10^-32 = (x)(8.5 x 10^-13 + 3x)^3

OH- concentration is very small

2 x 10^-32 = (x)(3x)^3

2 x 10^-32 = 27x^4

x = [OH-] = 5.22 x 10^-9 M

pOH = -log[OH-] = 8.28

pH = 14 - pOH = 5.72