A swimming pool was sufficiently alkaline so that the carbon dioxide absorbed from the air produced a solution in the pool that was 2 × 10^-4 M in carbonate ion. If the pool originally contained 4 × 10^-3 M Mg²⁺, 6 × 10^-4 M Ca²⁺, and 8 × 10^-7 M Fe²⁺, then the precipitate that was formed consisted of ________. [Ksp values are: CaCO₃, 4.7 × 10^-9; MgCO₃, 4.0 × 10^-5; FeCO₃, 2.0 × 10^-11]

Question

Quizplus · Accepted Answer

To determine which precipitates will form, we must calculate the ion product (IP) for each metal carbonate and compare it to their respective Ksp values.

IP = [Mg2+][CO32-] = (4 × 10^-3 M)(2 × 10^-4 M) = 8 × 10^-8

Ksp(MgCO3) = 4.0 × 10^-5

IP < Ksp, so MgCO3 will not precipitate.

IP = [Ca2+][CO32-] = (6 × 10^-4 M)(2 × 10^-4 M) = 1.2 × 10^-7

Ksp(CaCO3) = 4.7 × 10^-9

IP > Ksp, so CaCO3 will precipitate.

IP = [Fe2+][CO32-] = (8 × 10^-8 M)(2 × 10^-4 M) = 1.6 × 10^-11

Ksp(FeCO3) = 2.0 × 10^-11

IP < Ksp, so FeCO3 will not precipitate.

Therefore, the precipitate that will form consists of only CaCO3 and FeCO3, making choice D the correct answer.

A Swimming Pool Was Sufficiently Alkaline So That the Carbon