Question

The solubility of solid aluminum hydroxide (Al(OH)3) at 25∘C is 9.3×10−7gL. Determine the value of the solubility product (Ksp) for aluminum hydroxide.

Select the correct answer below:
5.5×10−31
7.5×10−25
2.1×10−32
4.8×10−28

Answer 1

A) 5.5E-31

Step-by-step explanation:

First, write the equation for the dissolution of aluminum hydroxide and the equation for the solubility product (Ksp)

Al(OH)3 (s) ⟷ Al3+ (aq) + 3OH− (aq) where Ksp = [Al3+][OH−]^3

To solve for Ksp, we need to calculate the molar solubility of Al(OH)3. The molar mass of aluminum hydroxide can be used to convert the given solubility in units of g/L into molar solubility in units of mol/L

9.3E−7 g/L × 1 mol Al(OH)3 / 78.003 g Al(OH)3 = 1.1923E−8 mol/L

Note that for every mole of Al(OH)3 that dissolves, 1 mol of Al3+ and 3 mol of OH− ions are produced

[Al3+] = 1.1923E−8 M

[OH−] = 3 × 1.1923E−8 M = 3.5768E−8 M

Therefore, Ksp can be calculated as follows

Ksp = [Al3+][OH−]^3 = (1.1923E−8)(3.5768E−8)^3 = 5.456E−31

Therefore, we find that Ksp is approximately equal to 5.5E−31