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Chapter 16: Problem 62

A solution contains $3.0 \times 10^{-3} M \mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2} .$ What concentrations of \(\mathrm{KF}\) will cause precipitation of solid \(\mathrm{MgF}_{2}\left(K_{\mathrm{sp}}=6.4 \times 10^{-9}\right) ?\)

Short Answer

Expert verified
The minimum concentration of \(\mathrm{KF}\) that will cause precipitation of solid \(\mathrm{MgF_2}\) is approximately \(1.46 \times 10^{-3} M\).

Step by step solution

01

1. Write the reaction equation

The equation for the formation of magnesium fluoride is given by: \[\mathrm{Mg}^{2+} + 2\mathrm{F}^- \rightleftharpoons \mathrm{MgF}_2\]
02

2. Write the Ksp expression for MgF2

According to the balanced chemical equation, the solubility product constant (Ksp) expression for magnesium fluoride (MgF2) is: \[K_{sp} = [\mathrm{Mg}^{2+}][\mathrm{F}^-]^2\] In this problem, the given value for Ksp is \(6.4 \times 10^{-9}\).
03

3. Determine the initial concentrations

The given concentration for \(\mathrm{Mg(NO_3)_2}\) is \(3.0 \times 10^{-3}M\). Since one mole of \(\mathrm{Mg(NO_3)_2}\) dissociates into one mole of \(\mathrm{Mg}^{2+}\), the initial concentration of \(\mathrm{Mg}^{2+}\) is also \(3.0 \times 10^{-3}M\). Let the concentration of \(\mathrm{KF}\) be denoted by x M. Since one mole of \(\mathrm{KF}\) dissociates into one mole of \(\mathrm{F}^-\), the initial concentration of \(\mathrm{F}^-\) is also x M.
04

4. Write the reaction quotient expression (Q)

The reaction quotient (Q) indicates the current state of the reaction and is given by the same expression as the solubility product constant (Ksp): \[Q = [\mathrm{Mg}^{2+}][\mathrm{F}^-]^2\] For our scenario, we will substitute the initial concentrations of \(\mathrm{Mg}^{2+}\) and \(\mathrm{F}^-\) into the expression for Q: \[Q = (3.0 \times 10^{-3})(x)^2\]
05

5. Determine the concentration of KF that will cause precipitation

Precipitation will occur when Q ≥ Ksp. Using the given Ksp value and our expression for Q, we have: \[(3.0 \times 10^{-3})(x)^2 \geq 6.4 \times 10^{-9}\] Now, solve for x: \[x^2 \geq \frac{6.4 \times 10^{-9}}{3.0 \times 10^{-3}}\] \[x^2 \geq 2.13 \times 10^{-6}\] \[x \geq \sqrt{2.13 \times 10^{-6}}\] \[x \geq 1.46 \times 10^{-3}M\] The minimum concentration of \(\mathrm{KF}\) that will cause precipitation of solid MgF2 is approximately \(1.46 \times 10^{-3} M\).

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Most popular questions from this chapter

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