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

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g