K_sp ; increasing pH with given [OH]

[lizzyb]

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.

         -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

         -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

         -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?

 

[Borek]

We're given [OH-] = 0.001 M ; plugging this into (1) gives [Mg2+] = 5.623e-6 M.

At this moment you have a correct answer almost ready. You know the final concentration of Mg in the solution, that's enough to calculate number of moles and mass dissolved in 1L. Whatever you did later didn't make any sense to me.

It will not be identical with the key, no idea why.

Note that concentration of 18.198e-6M is below the precipitation limit:

-log(18.198e-6 x 1e-3²) = 10.74, while you have listed pK_sp of 11.25.