Find molar solubility

Question

Determine the molar solubility for Cr(OH)3 by constructing an ICE table. Cr(OH)3 is dissolved in an aqueous solution that has a pH of 11.90 at 25 °C.

Answer 1

For this problem you would need the Ksp for Cr(OH)₃, which is 6.7 x 10^-31

Since the pH was given, you can find the initial concentration of [OH^-]

[OH^-] = 10^-pOH

pOH = 14 - pH

pOH = 14 - 11.90

pOH = 2.1

[OH-] = 10^-2.1

[OH-] = 7.94 x 10^-3M

For Ksp (molar solubility product), don't use solids or liquids, therefore you start your ICE table on the product side because they are in aqueous phase

Cr(OH)3(s) ↔ Cr³⁺(aq) + 3 OH-(aq)

I 0M 7.94 x 10^-3M

C +x +3x

E x 7.94 x 10^-3 + 3x

Since the Ksp is small, use the assumption rule. 7.94 x 10^-3M + 3(0) = 7.94 x10^-3M

Write the Ksp expression for this reaction and then solve for x

Ksp = [Cr³⁺][OH^-]³

6.7 x 10^-31 = [x][7.94 x 10^-3]³

6.7 x 10^-31 = [x] [5.0 x 10^-7]

6.7 x 10^-31 = x

5.0 x 10^-7

1.34 x 10^-24M = x (molar solubility of Cr(OH)₃)