Asked by Jamie
If the solubility of CaCO_3 in water at 25°C is 5.84 ✕ 10^−3 g/L, calculate the solubility-product constant (Ksp) of CaCO_3, assuming complete dissociation of the CaCO_3 that has dissolved.
Answers
Answered by
DrBob222
Convert 5.84E-3 g/L to mols/L.
mols/L = M = 5.84E-3/molar mass CaCO3 = ?
....................CaCO3 ==> Ca^2+ + [CO3]^2-
I....................solid..............0..............0
C...................solid..............x...............x
E....................solid..............x...............x
You know x = M from above. Plug that into the Ksp expression and solve for Ksp. Post your work if you get stuck.
mols/L = M = 5.84E-3/molar mass CaCO3 = ?
....................CaCO3 ==> Ca^2+ + [CO3]^2-
I....................solid..............0..............0
C...................solid..............x...............x
E....................solid..............x...............x
You know x = M from above. Plug that into the Ksp expression and solve for Ksp. Post your work if you get stuck.
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