Ce2ssierEchav
contestada

The solubility of lead (II) carbonate is 2.7x10^-7 mol/L. What is its ksp?
The correct answer is 7.3x10^-14, but I keep getting 7.87x10^-20

Respuesta :

Willchemistry
First, the solubility product of PbCO₃ :

PbCO₃ ⇆ Pb⁺²   +   CO₃⁻²

Ksp = [ Pb⁺²] [ CO₃⁻² ]

Kps = ( 2.7x10⁻⁷) *  ( 2,7x10⁻⁷)  ---> multiplique, add the exponents and conserve sign

Kps = 7.3x10⁻¹⁴

hope this helps!