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

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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