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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. 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.

Respuesta :

Answer: Option (b) is the correct answer.

Explanation:

The given data is as follows.

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

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

      [tex]K_{sp} (CaC_{2}O_{4}) = 2.3 \times 10^{-9}[/tex]

As expression for [tex]K_{sp}[/tex] will be as follows.

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

                            = [tex]2.3 \times 10^{-9}[/tex]

And, expression for ionic product will be as follows.

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

                      = [tex]2.1 \times 10^{-5} \times 4.75 \times 10^{-5}[/tex]

                      = [tex]9.975 \times 10^{-10}[/tex]

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.

When these solutions are mixed then,

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

Given:

Ksp ([tex]CaC_2O_4[/tex]) = [tex]2.3 *10^{-9}[/tex]

Calcium ion = [tex]2.1 * 10^{-5} M[/tex]

Oxalate ion = [tex]4.75 * 10^{-5} M[/tex]

Solubility product:

[tex]K_{sp}=[Ca^{2+}][C_2O_4^{2-}]=2.3*10^{-9}[/tex]

And, expression for ionic product will be as follows.

[tex]Q=[Ca^{2+}][C_2O_4^{2-}]\\\\Q=2.1*10^{-5}*4.75*10^{-5}\\\\Q=9.975*10^{-10}[/tex]

Thus,  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. So correct option is B.

Find more information about Solubility product here:

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