Question

Solubility product of CaC₂O₄ at a given temperature in pure water is 4×10^-9 (mol L^-1)². Solubility of CaC₂O₄ at the same temperature is:

• A. 6.3×10^-5 mol L^-1
• B. 2×10^-5 mol L^-1
• C. 2×10^-4 mol L^-1
• D. 6.3×10^-4 mol L^-1

Answer 1

The Correct Option is A

\(\text{CaC}_2\text{O}_4\) is a sparingly soluble salt that dissociates in water as follows:

\(\text{CaC}_2\text{O}_4(s) \rightleftharpoons \text{Ca}^{2+}(aq) + \text{C}_2\text{O}_4^{2-}(aq)\)

Let the solubility of \(\text{CaC}_2\text{O}_4\) be \(s\) \(\text{mol L}^{-1}\). Then, at equilibrium:

\([\text{Ca}^{2+}] = s\)

\([\text{C}_2\text{O}_4^{2-}] = s\)

The solubility product, \(K_{sp}\), is given by:

\(K_{sp} = [\text{Ca}^{2+}][\text{C}_2\text{O}_4^{2-}] = s \times s = s^2\)

Given that \(K_{sp} = 4 \times 10^{-9}\), we have:

\(s^2 = 4 \times 10^{-9}\)

\(s = \sqrt{4 \times 10^{-9}} = \sqrt{4} \times \sqrt{10^{-9}} = 2 \times 10^{-4.5} = 2 \times 10^{-4} \times \sqrt{0.1}=2 \times 10^{-4}\times 0.316\) Note that \(\sqrt{10^{-9}}=10^{-4.5}\) and \(10^{-0.5}=0.316\).

\(s= \sqrt{4\times10^{-9}} = 2\times 10^{-4.5} \text{ mol L}^{-1}\) Now \(2\times10^{-4.5} = 2\times \frac{1}{\sqrt{10^9}} = \frac{2}{10^{4.5}} = \frac{2}{10^4 \sqrt{10}} = \frac{2}{10^4 \times3.1622}= 6.32 \times 10^{-5}\)

Therefore the solubility of calcium oxalate will be:\(s= 6.32 \times 10^{-5}\)

\(s=6.32 *10^-5\) We calculate \(s= \sqrt{4}*10^{\frac{-9}{2}}\). If we take \(\sqrt{10}= 3.16\):\(10^{-4.5}=10^{-5}* \sqrt{10}\) or, \(s= 6.32 \times10^{-5}\) which is close to choice (A).

Final Answer: The final answer is A

Answer:

(A) \(6.3 \times 10^{-5} \text{ mol L}^{-1}\)

Answer 2

Write the Dissolution Equilibrium:

\(\text{CaC}_2\text{O}_4(s) \rightleftharpoons \text{Ca}^{2+}(aq) + \text{C}_2\text{O}_4^{2-}(aq)\)

Define Solubility (s):

Let \(s\) be the solubility of \(\text{CaC}_2\text{O}_4\) in \(\text{mol L}^{-1}\).

Express Ion Concentrations in terms of Solubility:

At equilibrium:

\([\text{Ca}^{2+}] = s\)

\([\text{C}_2\text{O}_4^{2-}] = s\)

Write the Expression for the Solubility Product (K_sp):

\(K_{sp} = [\text{Ca}^{2+}][\text{C}_2\text{O}_4^{2-}]\)

Substitute the Ion Concentrations into the K_sp Expression:

\(K_{sp} = s \times s = s^2\)

Solve for Solubility (s):

Given \(K_{sp} = 4 \times 10^{-9}\),

\(s^2 = 4 \times 10^{-9}\)

\(s = \sqrt{4 \times 10^{-9}}\)

\(s = \sqrt{4} \times \sqrt{10^{-9}}\)

\(s = 2 \times 10^{-4.5}\)

Convert the final equation into something easier to manage and solve:
\((s = 2 \times 10^{-4.5}\) can also be rewritten as follows:
\((s = 2 \times 10^{-5} \times \sqrt{10}\).
Since \(\sqrt{10}\)= 3.1622,
therefore, \(2 \times \sqrt{10}=6.3245\).
Therefore \(s = 6.3245 \times 10^{-5}\)
or \(s=6.3 \times 10^{-5}\).

Answer: (A) \(6.3 \times 10^{-5} \text{ mol L}^{-1}\)

Answer 3

Let the solubility of CaC₂O₄ be 's' mol L⁻¹. The dissociation of CaC₂O₄ in water can be represented as: \[ \text{CaC₂O₄} (s) \rightleftharpoons \text{Ca}^{2+} (aq) + \text{C₂O₄}^{2-} (aq) \] The solubility product (Ksp) is given by: \[ K_{sp} = [\text{Ca}^{2+}] [\text{C₂O₄}^{2-}] \] Since one mole of CaC₂O₄ produces one mole of Ca²⁺ and one mole of C₂O₄²⁻, we have: \[ K_{sp} = s \times s = s² \] We are given that the Ksp is 4 × 10⁻⁹ mol² L⁻². So, \[ s² = 4 \times 10^{-9} \] Solving for s: \[ s = \sqrt{4 \times 10^{-9}} = 2 \times 10^{-5} \, \text{mol L}^{-1} \] Thus, the solubility of CaC₂O₄ is 6.3 × 10⁻⁵ mol L⁻¹.