Question

# A 1.00-L solution saturated at$25 ^ { \circ } \mathrm { C }$with calcium oxalate$(CaC_2O_4)$contains 0.0061 g of$CaC_2O_4.$Calculate the solubility-product constant for this salt at$25 ^ { \circ } \mathrm { C }$

Solution

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

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

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

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

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

Volume of solution is $1\ \mathrm{dm^3}$ 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 $1:1$. Reaction is

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

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

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