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OpenStudy (anonymous):
How many liters of pure water would be required to dissolve 0.2500 grams of limestone, CaCO3, if the Ksp for calcium carbonate is 4.5 x 10-9.
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OpenStudy (anonymous):
Ksp= (Ca)(CO3)=4.9e-9 (Ca)=(CO3)=(CaCO3)=(sqrt(4.9e-9) mols CaCO3)/(L of H20) =.2500g/molar mass of CaCO3/xL cross multiply and solve for x first post hope its not too confusing
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