Question
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
The correct answer is 7.3x10^-14, but I keep getting 7.87x10^-20
Asked by: USER5129
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145 Answers
Answer (145)
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!
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!