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Ce2ssierEchav
Ce2ssierEchav Ce2ssierEchav
  • 16-03-2016
  • Chemistry
Answered

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

Verified Answer :

Willchemistry
Willchemistry Willchemistry
  • 16-03-2016

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!

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