## Related questions with answers

(a) Calculate the solubilit y of calcium oxalate (CaC2O4) in 1.0 M oxalic acid (H2C2O4) at 25°C, using the two acid ionization constants for oxalic acid and the solubility product Ksp = 2.6×10^-9 for CaC2O4. (b) Calculate the solubility of calcium oxalate in pure water at 25°C. (c) Account for the difference between the results of (a) and (b).

Solution

Verified$\textbf{a)}$Assume that $x$ M of CaC$_2$O$_4$ dissolves at equilibrium. The concentration of Ca$^{2+}$ denote by $x$ and C$_2$O$_4^{2-}$ reacts with H$_2$C$_2$O$_4$:

$\text{C$_2$O$_4^{2-}$} + \text{H$_2$C$_2$O$_4$} \rightleftharpoons 2 \text{HC$_2$CO$_4^-$}$

The $K$ is:

$K=\dfrac{[\text{HC$_2$CO$_4^-$}]^2}{[\text{H$_2$C$_2$O$_4$} ][\text{C$_2$O$_4^{2-}$}]}=\dfrac{K_{a_1}}{K_{a_2}}$

$K=\dfrac{6.5 \cdot 10^{-2}}{6.1 \cdot 10^{-5}}=1065$

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