Question

The Ksp of cobalt(II) hydroxide, `Co(OH)_2`, is `5.92 × 10^-15`. Calculate the molar solubility of this compound?

Answer 1

The molar solubility is `1.14 × 10^"-5"color(white)(l)"mol/L"`.

Explanation:

At equilibrium we have

`color(white)(mmmmmm)"Ca(OH)"_2 ⇌ "Ca"^"2+" + "2OH"^"-"`
`"E/mol·L"^"-1": color(white)(mmmmmmmll)xcolor(white)(mmm)2x`

`K_text(sp) = ["Ca"^"2+"]["OH"^"-"]^2 = x(2x)^2 = 4x^3 = 5.92 × 10^"-15"`

`x^3 = (5.92 × 10^"-15")/4 = 1.48 ×10^"-15"`

`x = 1.14 × 10^"-5"`

The molar solubility is `1.14 × 10^"-5"color(white)(l)"mol/L"`.