A solution contains 0.10 M Ca²⁺ and 0.10 M Mg²⁺.The pH of the solution is raised without changing the volume of the solution.What percentage of Mg²⁺ remains in solution when Ca(OH)₂(s)first begins to precipitate? (K_sp of Ca(OH)₂ = 4.0 × 10−⁶ and K_sp of Mg(OH)₂ = 7.1 × 10−¹²)

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To find the percentage of Mg(OH)2 that remains in solution when Ca(OH)2 begins to precipitate, we first need to determine the pH at which Ca(OH)2 starts to precipitate. This is when the ion product of Ca2+ and OH- equals the Ksp of Ca(OH)2. Given Ksp of Ca(OH)2 = 4.0 × 10^-6, and assuming the concentration of Ca2+ is approximately 0.10 M at the start of precipitation, we can solve for [OH-] using the Ksp expression: Ksp = [Ca2+][OH-]^2. Solving for [OH-] gives a concentration that leads to a pH where Ca(OH)2 precipitates. At this pH, we can calculate the solubility of Mg(OH)2 using its Ksp (7.1 × 10^-12) and the [OH-] concentration. Since Mg(OH)2 has a lower Ksp than Ca(OH)2, it is more soluble in water and will not precipitate until a higher pH. The percentage of Mg(OH)2 remaining in solution can be calculated by comparing the initial concentration of Mg2+ to the concentration of Mg2+ at the point where Ca(OH)2 begins to precipitate, using the ion product of Mg2+ and OH- at that pH. Given the Ksp values and the process described, the correct calculation leads to the conclusion that a very small percentage of Mg(OH)2 remains in solution, which corresponds to choice A, 1.8 × 10^-14%. This calculation involves understanding the solubility product constant (Ksp) and how it relates to the solubility of ionic compounds in solution, particularly in relation to changes in pH.

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