Question

A solution containing CaCl2 is mixed with a solution of Li2C2O4 to form a solution that is 3.5 × 10-4 M in calcium ion and 2.33 × 10-4 M in oxalate ion. What will happen once these solutions are mixed? Ksp (CaC2O4) = 2.3 × 10-9.

Answer 1

This is an incomplete question, here is a complete question.

A solution containing CaCl₂ is mixed with a solution of Li₂C₂O₄ to form a solution that is 3.5 × 10⁻⁴ M in calcium ion and 2.33 × 10⁻⁴ M in oxalate ion. What will happen once these solutions are mixed? Ksp (CaC₂O₄) = 2.3 × 10⁻⁹

A. Nothing will happen since both calcium chloride and lithium oxalate are soluble compounds.

B. Nothing will happen Ksp > Q for all possible precipitants.

C. A precipitate will form as calcium oxalate is not soluble to any extent.

D. A precipitate will form since Q > Ksp for calcium oxalate.

Answer : The correct option is, (D) A precipitate will form since Q > Ksp for calcium oxalate.

Explanation :

Concentration of $Ca^{2+}$ = $3.5\times 10^{-4}M$

Concentration of $C_2O_4^{2-}$ = $2.33\times 10^{-4}M$

The equilibrium chemical reaction will be:

$CaC_2O_4\rightleftharpoons Ca^{2+}+C_2O_4^{2-}$

The solubility constant expression for this reaction is:

$K_{sp}=[Ca^{2+}][C_2O_4^{2-}]=2.3\times 10^{-9}$

Now we have to calculate the ionic product for calcium oxalate.

$Q_{sp}=[Ca^{2+}][C_2O_4^{2-}]$

$Q_{sp}=(3.5\times 10^{-4})\times (2.33\times 10^{-4)$

$Q_{sp}=8.2\times 10^{-8}$

From this we conclude that, $Q_{sp}>K_{sp}$ that means a white solid precipitate of calcium oxalate will be formed when the solutions are mixed.

Hence, the correct option is, (D)