Answer to Question #325687 in General Chemistry for Tlove

Question

The molar solubility of aluminum hydroxide in a 0.179 M aluminum acetate solution is

M.

Answer 1

Al(CH₃COO)₃ = Al³⁺ + 3CH₃COO^-

[Al³⁺] = 0.179 mol

K_sp = [Al³⁺] × [OH^-]³ = (0.179)(3S)³ = 4.833S³

S³ = K_sp / 4.833 = 1.3 × 10^-33 / 4.833 = 2.7 × 10^-32

S = 6.5 × 10^-12 mol/L