bradleycawley02
24.08.2019 •
Chemistry
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
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Ответ:
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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