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vivado [14]
4 years ago
9

Mg(OH)2 is a sparingly soluble compound, in this case a base, with a solubility product, Ksp, of 5.61×10−11. It is used to contr

ol the pH and provide nutrients in the biological (microbial) treatment of municipal wastewater streams.
Based on the given value of the Ksp, what is the molar solubility of Mg(OH)2 in pure H2O?
Chemistry
1 answer:
VLD [36.1K]4 years ago
4 0

Answer:

1.27 x 10⁻⁴M

Explanation:

For a 2 to 1 ionization ratio, solubility in pure water can be calculated using the formula S = ∛(Ksp/27) = ∛(5.61 x 10⁻¹¹/27 = 1.27 x 10⁻⁴M.

Mg(OH)₂ ⇄ Mg⁺² + 2OH⁻ => Ksp = [Mg⁺²][OH⁻]² = (x)(2x)² = 4x³

Solve for 'x' => x = Solubility = ∛Ksp/4

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The question is incomplete, here is the complete question.

A chemist prepares a solution of copper(II) fluoride by measuring out 0.0498 g of copper(II) fluoride into a 100.0mL volumetric flask and filling the flask to the mark with water.

Calculate the concentration in mol/L of the chemist's copper(II) fluoride solution. Round your answer to 3 significant digits.

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To calculate the molarity of solute, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

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Putting values in above equation, we get:

\text{Molarity of copper (II) fluoride)=\frac{0.0498\times 1000}{101.54\times 100.0}\\\\\text{Molarity of copper (II) fluoride}=4.90\times 10^{-3}mol/L

Hence, the concentration of copper fluoride in the solution is 4.90\times 10^{-3}mol/L

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