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

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SUMMARY

The discussion focuses on calculating the hydronium ion concentration in a magnesium hydroxide (Mg(OH)2) solution. A participant initially calculated the hydroxide ion concentration as [OH-] = 2.058 * 10^-3 M, leading to a pH of 11.31 and a hydronium ion concentration of [H3O+] = 5 * 10^-12 M. However, through further calculations, the correct hydroxide concentration was determined to be [OH-] = 4.11 * 10^-3 M, resulting in a pH of 11.9 and a hydronium ion concentration of [H3O+] = 1.2 * 10^-12 M. The final values reflect the accurate stoichiometry of magnesium hydroxide dissociation.

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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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