Question

(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

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a)\textbf{a)}Assume that xx M of CaC2_2O4_4 dissolves at equilibrium. The concentration of Ca2+^{2+} denote by xx and C2_2O42_4^{2-} reacts with H2_2C2_2O4_4:

C2O42+H2C2O42HC2CO4\text{C$_2$O$_4^{2-}$} + \text{H$_2$C$_2$O$_4$} \rightleftharpoons 2 \text{HC$_2$CO$_4^-$}

The KK is:

K=[HC2CO4]2[H2C2O4][C2O42]=Ka1Ka2K=\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=6.51026.1105=1065K=\dfrac{6.5 \cdot 10^{-2}}{6.1 \cdot 10^{-5}}=1065

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