What is the Hydronium Ion Concentration in a Magnesium Hydroxide Solution?

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The discussion focuses on calculating the hydronium ion concentration in a magnesium hydroxide solution. Initially, the participant calculated the hydroxide ion concentration as 2.058 x 10^-3 M but later corrected it to 4.11 x 10^-3 M after realizing the stoichiometry of magnesium hydroxide dissociation. The pOH was recalculated to be 2.08, leading to a pH of 11.9. Consequently, the hydronium ion concentration was found to be 1.2 x 10^-12 M. The conversation emphasizes the importance of accurately determining molar concentrations in chemical calculations.
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Homework Statement



If 0.12 g of magnesium hydroxide Mg(OH)2 is dissolved in sufficient water to make 500. mL of solution, what is the hydronium ion concentration in the solution? (Assume that all of the magnesium hydroxide dissolves.)

Homework Equations



pH = -log [OH-]
pOH = -log [H3O+]

The Attempt at a Solution



I got 5 * 10^-12:

Mg(OH)2 --> Mg2+ + 2OH-
[OH-] = 2.058 * 10^-3 M

pOH = 2.6852
then pH = 14 - pOH = 11.31

[H3O+] = 10^-11.30
= 5*10^-12

is this the right method, is there an easier way?
 
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Approach is correct, but the result is wrong. What is molar concentration of Mg(OH)2?
 
Last edited:
M = .12/.5 = .24
 
got it.. thanks though
 
moles Mg(OH)2 = 2.058*10^-3

Mg(OH)2 --> Mg2+ + 2OH-
1 mol Mg(OH)2 ---> 2 mol OH-
2.058*10^-3 mol Mg(OH)2---> 2 * 2.058*10^-3 = 4.11*10^-3


thus [OH-] = 4.11*10^-3 M
pOH = 2.08
then pH = 14 - pOH = 11.9

[H3O+] = 10^-11.9
= 1.2*10^-12

:) is this rite?
 
a.a said:
2.058*10^-3 mol Mg(OH)2---> 2 * 2.058*10^-3 = 4.11*10^-3

This is number of moles, not concentration.
 
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