Question

A 1.00-L solution saturated at

25C25 ^ { \circ } \mathrm { C }

with calcium oxalate

(CaC2O4)(CaC_2O_4)

contains 0.0061 g of

CaC2O4.CaC_2O_4.

Calculate the solubility-product constant for this salt at

25C25 ^ { \circ } \mathrm { C }

Solution

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Solubility product is product of concentrations of ions in saturated solution. Mass of CaC2O4\mathrm{CaC_2O_4} is 0.0061 g0.0061\ \mathrm{g}. We use formula

n=mMn=\dfrac{m}{M}

where m is mass, n is number of moles and M is molar mass we find n.

M(CaC2O4)=Ar(Ca)+2Ar(C)+4Ar(O)=128 gmol\mathrm{M}\left(\mathrm{CaC_2O_4}\right)=\mathrm{Ar}\left(\mathrm{Ca}\right)+2\mathrm{Ar}\left(\mathrm{C}\right)+4\mathrm{Ar}\left(\mathrm{O}\right)=128\ \frac{\mathrm{g}}{\mathrm{mol}}

n=4.76105 mol\mathrm{n}=4.76\cdot 10^{-5}\ \mathrm{mol}

Volume of solution is 1 dm31\ \mathrm{dm^3} so concentration is equal to number of moles and because solution is saturated this is equal to concentration of calcium and oxalate ion, because mol ratio 1:11:1. Reaction is

CaC2O4Ca2++C2O42\mathrm{CaC_2O_4 \rightarrow Ca^{2+} + C_2O_4^{2-}}

Ksp=(Ca2+)(C2O42)=4.761054.76105=22.651010 mol2dm6\mathrm{K}_{\mathrm{sp}}=\:\left(\mathrm{Ca^{2+}}\right)\left(\mathrm{C_2O_4^{2-}}\right)=4.76\cdot 10^{-5}\cdot 4.76\cdot 10^{-5}=22.65\cdot 10^{-10}\ \frac{\mathrm{mol^2}}{\mathrm{dm^6}}

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