1. State the Problem A water initially contains 40 mg/L of Mg2+. The pH of the water is increased until the concentration of hydroxide ions (OH-) is 0.001000 M. What is the concentration of magnesium ion in this water at this pH? Give you answer in milligrams per liter. Assume that the temperature of the solution is 25oC. 2. Relevant Material Mg(OH)2 (s) <---> Mg2+ + 2 OH- pKs = 11.25 Answer (given in book): 0.4423 mg / L 3. Work Done So Far The initial molarity is 1.646e-3 M. The given answer molarity is 18.198e-6 M. Code (Text): -11.25 2 K_sp = 10 = [Mg2+] [OH-] (1) We're given [OH-] = 0.001 M ; plugging this into (1) gives [Mg2+] = 5.623e-6 M. let [Mg2+] = s so [OH-] = 2s Then Code (Text): -11.25 2 3 K_sp = 10 = s * (2s) = 4 s (2) so s = 1.12e-4 Finally, I tried setting up a similar equation as (2): let [Mg2+] = 1.646e-3 - s, [OH-] = s, then Code (Text): -11.25 2 K_sp = 10 = (1.646e-3 - s) * (2s) (3) so s = 2.949e-5 (the final [Mg2+] = 1.646e-3 - s = 1.6165e-3 M What am I not doing correctly???