Question

× 10-5 M in oxalate ion. What will happen once these solutions are mixed? Ksp (CaC2O4) = 2.3 × 10-9. A solution containing CaCl2 is mixed with a solution of Li2C2O4 to form a solution that is 2.1 × 10-5 M in calcium ion and 4.75 × 10-5 M in oxalate ion. What will happen once these solutions are mixed? Ksp (CaC2O4) = 2.3 × 10-9. Nothing will happen since both calcium chloride and lithium oxalate are soluble compounds. Nothing will happen since Ksp > Q for all possible precipitants. A precipitate will form since Q > Ksp for calcium oxalate. Nothing will happen since calcium oxalate is extremely soluble. There is not enough information to determine.

Answer 1

Answer: Option (b) is the correct answer.

Explanation:

The given data is as follows.

      $[Ca^{2+}] = 2.1 \times 10^{-5}$ M

      $[C_{2}O_{4}^{2-}] = 4.75 \times 10^{-5}$ M

      $K_{sp} (CaC_{2}O_{4}) = 2.3 \times 10^{-9}$

As expression for $K_{sp}$ will be as follows.

                  $K_{sp} = [Ca^{2+}][C_{2}O^{2-}_{4}]$

                            = $2.3 \times 10^{-9}$

And, expression for ionic product will be as follows.

                  Q = $[Ca^{2+}][C_{2}O^{2-}_{4}]$

                      = $2.1 \times 10^{-5} \times 4.75 \times 10^{-5}$

                      = $9.975 \times 10^{-10}$

Since, it is shown here that ionic product is less that solubility product. Hence, no precipitate will form.

Thus, we can conclude that nothing will happen since Ksp > Q for all possible precipitants.