Question

The solubility product constant for `"Mg(OH)"_2` is `1.8 × 10^"-11"`. What would be the solubility of `"Mg(OH)"_2` in 0.345 M `"NaOH"`?

The solubility product constant for `"Mg(OH)"_2` is `1.8 × 10^"-11"`. What would be the solubility of `"Mg(OH)"_2` in 0.345 M `"NaOH"`?

Answer 1

The solubility would be `1.5 × 10^"-10"color(white)(l) "mol/L"`.

Explanation:

Set up the chemical equation for the equilibrium:

`color(white)(mmmmmm)"Mg(OH)"_2"(s)" ⇌ "Mg"^(2+)"(aq)" + 2"OH"^"-""(aq)"`
`"E:/mol·L"^"-1" color(white)(mmmmmmmmmm)x color(white)(mmmm)0.345 + 2x`

Set up the solubility product expression:

`K_"sp" = ["Mg"^(2+)]["OH"^"-"]^2 = x(0.345 +2x)^2 = 1.8 × 10^"-11"`

Check that `2xcolor(white)(l) "<< 0.345"`

`0.345/(1.8 × 10^"-11") = 1.9 × 10^10 > 400`. ∴ `2xcolor(white)(l) "<< 0.345"`,

Solve the `K_"sp"` expression:

`x(0.345)^2 = 1.8 × 10^"-11"`

`0.119x = 1.8 × 10^"-11"`

`x = (1.8 × 10^"-11")/0.119 = 1.5 × 10^"-10"`

∴ Solubility of #"Mg(OH)"_2 = 1.5 × 10^"-10" color(white)(l) "mol/L"#