Final answer:
To calculate the solubility of aluminum hydroxide at pH 11, we use the Ksp value and the Henderson-Hasselbalch equation to find the OH- concentration, then solve for Al3+ concentration. The molar solubility is found to be 2 × 10^-23 M, which does not match the provided options, suggesting an error in the question's provided answers.
Step-by-step explanation:
The solubility of aluminum hydroxide, Al(OH)3, in a solution buffered at pH 11 can be calculated using the solubility product constant (Ksp) and the Henderson-Hasselbalch equation to find the concentration of hydroxide ions (OH-) in the buffer. The Ksp for Al(OH)3 is 2 × 10-32. At pH 11, the pOH is equal to 3 (since pH + pOH = 14), and the concentration of OH- can be calculated as 10-3 M.
Al(OH)3 will dissociate as follows: Al(OH)3 → Al3+ + 3OH-. The solubility product expression is Ksp = [Al3+]×[OH-]3. Substituting the concentration of OH- and solving for [Al3+] gives us the molar solubility.
Therefore, the molar solubility of Al(OH)3 in a solution buffered at pH 11 is [Al3+] = (Ksp/[OH-]3) which is (2 × 10-32)/(10-3)3 = 2 × 10-23 M, which is not one of the provided options, indicating a possible error in the question or the answer choices. However, the methodology described is how one would approach such a calculation.