- #1
lizzyb
- 168
- 0
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.
We're given [OH-] = 0.001 M ; plugging this into (1) gives [Mg2+] = 5.623e-6 M.
let [Mg2+] = s so [OH-] = 2s
Then
so s = 1.12e-4
Finally, I tried setting up a similar equation as (2):
let [Mg2+] = 1.646e-3 - s, [OH-] = s, then
so s = 2.949e-5 (the final [Mg2+] = 1.646e-3 - s = 1.6165e-3 M
What am I not doing correctly?
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:
-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:
-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:
-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?