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Chapter 16: Problem 89

Which cation will precipitate first as a carbonate mineral from an equimolar solution of \(\mathrm{Mg}^{2+}, \mathrm{Ca}^{2+},\) and \(\mathrm{Sr}^{2+} ?\)

Short Answer

Expert verified
Answer: Sr虏鈦� will precipitate first as a carbonate mineral because strontium carbonate (SrCO鈧�) has the smallest solubility product constant among the three carbonate minerals.

Step by step solution

01

Write the balanced chemical equations for the formation of the carbonate minerals

Write the balanced double displacement reaction equations for the formation of carbonate minerals from the given cations and the \(\mathrm{CO}_{3}^{2-}\) ion. Mg\({}^{2+}\) + CO\(_3{}^{2-}\) \(\to\) MgCO\(_3\) (Magnesium carbonate) Ca\({}^{2+}\) + CO\(_3{}^{2-}\) \(\to\) CaCO\(_3\) (Calcium carbonate) Sr\({}^{2+}\) + CO\(_3{}^{2-}\) \(\to\) SrCO\(_3\) (Strontium carbonate)
02

Identify the solubility products for each carbonate mineral

The solubility product constant \((K_{sp})\) for each carbonate mineral is a measure of how soluble the compound is in water. The smaller the solubility product, the less soluble the compound, and the more likely it is to precipitate. Look up the solubility product constants for magnesium carbonate, calcium carbonate, and strontium carbonate. MgCO\(_3\): \(K_{sp} = 3.5 \times 10^{-8}\) CaCO\(_3\): \(K_{sp} = 3.3 \times 10^{-9}\) SrCO\(_3\): \(K_{sp} = 5.6 \times 10^{-10}\)
03

Compare the solubility products to determine which cation will precipitate first

Now, compare the solubility product constants for the carbonate minerals of each given cation. The cation with the lowest solubility product will precipitate first as a carbonate mineral. In this case, the order of the solubility product constants is: SrCO\(_3 < \) CaCO\(_3 < \) MgCO\(_3\) Since the solubility product constant of strontium carbonate \((SrCO_3)\) is the smallest, this means that \(\mathrm{Sr}^{2+}\) will precipitate first as a carbonate mineral from an equimolar solution of \(\mathrm{Mg}^{2+}, \mathrm{Ca}^{2+},\) and \(\mathrm{Sr}^{2+}\).

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