Calcium oxalate \(\left(\mathrm{CaC}_{2} \mathrm{O}_{4}\right)\) is relatively insoluble in water \(\left(K_{\mathrm{sp}}=2 \times 10^{-9}\right) .\) However, calcium oxalate is more soluble in acidic solution. How much more soluble is calcium oxalate in 0.10\(M \mathrm{H}^{+}\) than in pure water? In pure water, ignore the basic properties of \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}.\)

Short Answer

Expert verified
The solubility of calcium oxalate in pure water is \(1.41 \times 10^{-5} M\), calculated using the given \(K_{sp}\) value. The solubility in an acidic solution would be higher, as the oxalate ion reacts with the H+ ions. However, without more information about the solubility constant in acidic conditions (\(K_{solubility}\)), it is not possible to determine the exact difference in solubilities.

Step by step solution

01

Find the solubility of calcium oxalate in water

In water, the solubility equilibrium for calcium oxalate can be represented as: \[CaC_{2}O_{4}(s) \rightleftharpoons Ca^{2+}(aq) + C_{2}O_{4}^{2-}(aq)\] Let the solubility of calcium oxalate be represented by 's'. At equilibrium: \[K_{sp} = [Ca^{2+}][C_{2}O_{4}^{2-}] = s * s = s^2\] Given that, \(K_{sp}=2 \times 10^{-9}\) Now, solving for s: \[s = \sqrt{2 \times 10^{-9}}\] \[s = 1.41 \times 10^{-5}\: M\] The solubility of calcium oxalate in pure water is 1.41 x 10^-5 M.
02

Determine the solubility of calcium oxalate in a 0.10 M H+ solution

In an acidic solution, the oxalate ion will react with the H+ ions: \[H^+ + C_{2}O_{4}^{2-} \rightleftharpoons HC_{2}O_{4}^-\] In 0.10 M H+ solution, the concentration of H+ ions will be much greater than the concentration of C2O4^2- ions, and the reaction will shift towards the products, which will increase the solubility of calcium oxalate. Now, let the solubility of calcium oxalate in the acidic solution be represented by 's_acid'. The concentration of H+ ions will be 0.10 M. At equilibrium: \[ s_{acid}[\mathrm{H^+}] = K_{solubility}\] \[ K_{solubility} = \frac{s_{acid} * 0.10\: M}{s}\] \[ s_{acid} = \frac{K_{solubility} * s}{0.10\: M}\]
03

Calculate the difference in solubilities

To find how much more soluble calcium oxalate is in the acidic solution compared to pure water, subtract the solubility in water from the solubility in the acidic solution: \[Difference = s_{acid} - s = \frac{K_{solubility} * s}{0.10\: M} - s\] However, we are not given the K_solubility value. In this case, we can't solve the problem further without that information or any additional information.

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