A 1.00-L solution saturated at 25 C with calcium oxalate CaC2O4 contains 0.0061 g of CaC2O4.
Calculate the solubility-product constant for this salt at 25 C.
2 Answers
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CaC2O4 (s) ↔ Ca++ (aq) + C2O4– (aq)
Ksp = [Ca++] [C2O4–]
The molecular mass of CaC2O4 is 128.1 g/mole so the concentration of the saturated solution is:
(0.0061 g) / (1.00 L) (128.1 g/mol) = 4.76×10^-5 M.
Since one Ca++ ion and one C2O4– ion dissolve for each CaC2O4 that dissolves,
Ksp = [4.76×10^-5] [4.76×10^-5] = 2.27×10^-9.
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good but your lack of full use of divide signs makes this confusing just one of these
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and it would of been fine
