Magnesium hydroxide is used in several antacid formulations. when it is added to water it dissociates into magnesium and hydroxide ions. mg(oh)2(s) mg2+(aq) + 2oh-(aq) the equilibrium constant at 25°c is 8.9 ´ 10-12. one hundred grams of magnesium hydroxide is added to 1.00 l of water and equilibrium is established. what happens to the solution if another 10 grams of mg(oh)2 are now added to the mixture?

If you add 10 more grams of Mg(OH)₂, the solution will shift the equilibrium to the ions production to consume the excessive Mg(OH)₂.

Explanation:

Mg(OH)₂(s) → Mg²⁺(aq) + 2OH⁻(aq)

K = 8.9 x 10⁻¹²

K = [Mg²⁺][OH⁻]²

The concentration of Mg(OH)₂ does not enter the equation because it is in the solid state.

[Mg²⁺] = x

[OH⁻]²  = 2x

K = x.(2x)²

K = 4x³

4x³ =  8.9 x 10⁻¹²

x = 2.07 x 10⁻⁴ mol/L

At equilibrium:

[Mg²⁺] = x = 2.07 x 10⁻⁴ mol/L

[OH⁻]²  = 2x = 4.14 x 10⁻⁴ mol/L

i believe it is a.) released into

explanation:

you gotta add the following molecules so we can give u an answer

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