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If we have a solution that contains 0.022 M Calcium and 0.014 M Magnesium and we have to use...

Question:

If we have a solution that contains 0.022 M Calcium and 0.014 M Magnesium and we have to use potassium hydroxide to selectively precipitate one of the ions, while leaving the other ion in the solution.

A) Which metal will precipitate first? ( this metal will be referred to as " first ion")

B) What minimum concentration of KOH is necessary to begin precipitation of "first ion"?

C) What molarity of "first ion" remains in solution at the point when the second ion begins to precipitate?

D) As we set up the second trial for this same experiment, we realized that we ran out of potassium hydroxide. All we have in the storage cabinet is potassium carbonate. Would this be a good titrant to use for the selective precipitation? Why or why not? (Ksp of Mg(OH){eq}_2 {/eq} = 6.3 x 10{eq}^{-10} {/eq}, Ksp of Ca(OH){eq}_2 {/eq} = 6.5 x 10{eq}^{-6} {/eq})

Selective Precipitation:

The process by which any specific reagent would precipitate a given ion among a mixture of ions in an aqueous solution based on their solubility product is known as the selective precipitation of ions.

Answer and Explanation: 1

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Given Data

{eq}{\rm K_{sp}}\; {/eq} of {eq}\;\rm Mg{\left( {\rm OH} \right)_2} {/eq} = {eq}6.3 \times {10^{ - 10}} {/eq}

{eq}{\rm...

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The Common Ion Effect and Selective Precipitation

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Chapter 11 / Lesson 6
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Learn about the common ion effect on solubility. Learn what a common ion is, examine a common ion chart, and work through examples of the common ion effect.


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