The magnesium and calcium ions present in seawater ([Mg2+] = 0.059 M and [Ca2+]

Question

The magnesium and calcium ions present in seawater ([Mg2+] = 0.059 M and [Ca2+] = 0.011M) ca be separated by selective preciptiation with KOH.

A. What minimum [OH] triggers the preciptation of Mg2+ ion? Answer was 1.9 x 10^-6 M

B: You add KOH to the solution in part a. When the [OH] reaches 1.9 x 10^-6 as was calculated, magnesium hydroxide begins to precipitate out of solution. As you continue to add KOH, the magnesium hydroxide continues to precipitate. However, at some point, the [OH] becomes high enough to begin to precipatate the calcium ions as well. What is the concentration of Mg2+ when Ca2+ begins to precipitate? Answer was 4.9x 10^-10 M

C. Have you succesfully separated the calcium ions from the magnesium ions. Explain/show proof.

I am a little stumped on this question. especially letter C. Could you please show the step by step process with the work.

Explanation / Answer

Ksp(Mg(OH)2) = 2.06 x 10-13

Ksp(Ca(OH)2) = 4.68 x 10-6

Q = [Mg2+ ][OH- ]2

= (0.059)[OH-]2

Q= Ksp                          saturated solution and precipitation just begins.

2.06 x 10-13 = (0.059)[OH-]2

[OH-]2 = 1.9 x 10-6M

Ksp(Mg(OH)2) = 2.06 x 10-13

Ksp(Ca(OH)2) = 4.68 x 10-6

When Ca 2+ begins to precipitate [ OH- ]

Q = [Ca2+ ][OH- ]2

(0.011) [OH- ]2 = Ksp(Ca(OH)2)

(0.011) [OH- ]2 = 4.68 x 10-6

[OH- ]2 = 2.06 x 10-2M

Ksp(Mg(OH)2) = 2.06 x 10-13

Ksp(Ca(OH)2) = 4.68 x 10-6

[OH- ]2 = 2.06 x 10-2 M

When Ca2+ just begins to precipitate.

Now when Ca2+ begins to precipitate, Mg2+

Ksp = [Mg2+ ][OH- ]2

2.06 x 10-13 = [Mg2+](2.06 x 10-2)]2

[Mg2+ ] = 2.06 x 10-13 /2.06 x 10-2

= 4.9 x 10 -10M

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