The K_sp of aluminum hydroxide,Al(OH)₃,is 2 × 10−³¹.What pH is required to limit the Al³⁺ concentration to less than or equal to 1 × 10−¹⁰ M?

Question

Quizplus · Accepted Answer

The equilibrium equation for the dissolution of aluminum hydroxide is: Al(OH)3 (s) ⇌ Al3+ (aq) + 3 OH- (aq) The solubility constant, K_sp, for this reaction is 2 × 10−³¹, which means that at equilibrium, the product of the concentrations of Al3+ and OH- ions is equal to 2 × 10−³¹. Let's assume that the initial concentration of Al3+ is zero, so the initial concentration of OH- is also zero. As aluminum hydroxide dissolves, the concentration of OH- ions increases, and Al3+ ions are produced in the process. At equilibrium, the concentration of Al3+ ions will be equal to the concentration of OH- ions produced. Let x be the concentration of OH- ions produced at equilibrium. Then the concentration of Al3+ ions will also be x, and we can write: Ksp = [Al3+][OH-]3 2 × 10−³¹ = x * x3 x4 = 2 × 10−³¹ x = (2 × 10−³¹)1/4 x = 6.31 × 10−³ The pH of a solution is defined as the negative logarithm of the concentration of H+ ions. In this case, we want to limit the concentration of Al3+ ions to less than or equal to 1 × 10−¹⁰ M, which means that the concentration of OH- ions should be at least 1 × 10−¹⁰ M. The concentration of H+ ions required to produce this concentration of OH- ions can be calculated from the following equation: Kw = [H+][OH-] 1.0 × 10−⁴¹ = x * 1.0 × 10−¹⁰ x = 1.0 × 10−⁴³ The pH required to limit the concentration of Al3+ ions to less than or equal to 1 × 10−¹⁰ M is: pH = -log[H+] pH = -log(1.0 × 10−⁴³) pH = 3.6 Therefore, the correct answer is choice (C).

The Ksp of Aluminum Hydroxide,Al(OH)3,is 2 × 10−31

The K_sp of Aluminum Hydroxide,Al(OH)₃,is 2 × 10−³¹