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Chapter 17: Problem 48

The fluoride ion concentration in a saturated solution of magnesium fluoride, \(\mathrm{MgF}_{2}\), is \(2.3 \times 10^{-3} \mathrm{M}\). Calculate the value for the equilibrium constant.

Short Answer

Expert verified
The equilibrium constant \( K_{sp} \) is approximately \( 6.08 \times 10^{-9} \).

Step by step solution

01

Write the Dissociation Equation

The dissociation of magnesium fluoride in water can be represented by the equation: \[ \mathrm{MgF}_{2 (s)} \rightleftharpoons \mathrm{Mg}^{2+}_{(aq)} + 2 \mathrm{F}^-_{(aq)} \] This equation shows that one molecule of \( \mathrm{MgF}_2 \) dissociates into one magnesium ion and two fluoride ions.
02

Determine Ion Concentrations

According to the equation, for every 1 mole of \( \mathrm{MgF}_2 \) that dissolves, 1 mole of \( \mathrm{Mg}^{2+} \) ions and 2 moles of \( \mathrm{F}^- \) ions are produced. Given that the fluoride ion concentration \([\mathrm{F}^-]\) is \(2.3 \times 10^{-3} \ \mathrm{M}\), the magnesium ion concentration \([\mathrm{Mg}^{2+}]\) must be \(\frac{1}{2} \times 2.3 \times 10^{-3} \ \mathrm{M}\ = 1.15 \times 10^{-3} \ \mathrm{M}\).
03

Write the Expression for the Solubility Product Constant

The solubility product constant \( K_{sp} \) for the dissociation can be expressed as: \[ K_{sp} = [\mathrm{Mg}^{2+}][\mathrm{F}^-]^2 \] This expression captures the relationship between the ion concentrations at equilibrium.
04

Insert the Ion Concentrations into the Expression

Substitute the known concentrations into the \( K_{sp} \) expression: \[ K_{sp} = (1.15 \times 10^{-3}) \times (2.3 \times 10^{-3})^2 \] Calculate this step by step: \((2.3 \times 10^{-3})^2 = 5.29 \times 10^{-6}\), and then \( K_{sp} = (1.15 \times 10^{-3}) \times 5.29 \times 10^{-6}\).
05

Calculate the Solubility Product Constant

Complete the multiplication to find \( K_{sp} \): \[ K_{sp} = 1.15 \times 5.29 \times 10^{-9} = 6.0835 \times 10^{-9} \] So, the equilibrium constant \( K_{sp} \) is approximately \( 6.08 \times 10^{-9} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Dissociation Equation
When a compound dissolves in water, it breaks apart into its individual ions. This process is called dissociation. For magnesium fluoride \( \mathrm{MgF}_2 \), when it dissolves in water, it splits into magnesium ions \( \mathrm{Mg}^{2+} \) and fluoride ions \( \mathrm{F}^- \). The dissociation equation is written as: \[ \mathrm{MgF}_{2 (s)} \rightleftharpoons \mathrm{Mg}^{2+}_{(aq)} + 2 \mathrm{F}^-_{(aq)} \] This equation tells us the ratio in which ions are formed:
  • 1 molecule of \( \mathrm{MgF}_2 \) produces 1 \( \mathrm{Mg}^{2+} \) ion.
  • 1 molecule of \( \mathrm{MgF}_2 \) produces 2 \( \mathrm{F}^- \) ions.
Understanding this ratio is important for calculating ion concentrations in a solution.
Solubility Product Constant
The Solubility Product Constant, or \( K_{sp} \), is a specific type of equilibrium constant that applies to the solubility of ionic compounds. It tells us how much compound can dissolve in a solution to form its respective ions. For our case with \( \mathrm{MgF}_2 \), the expression for \( K_{sp} \) is given by: \[ K_{sp} = [\mathrm{Mg}^{2+}][\mathrm{F}^-]^2 \] This expression reflects how the solubility product relies on the concentrations of the ions in the solution. The exponent 2 on \( [\mathrm{F}^-] \) comes from two fluoride ions being produced from one formula unit of \( \mathrm{MgF}_2 \). Knowing \( K_{sp} \) helps to predict whether a precipitate will form or dissolve in a given solution.
Ion Concentration
Ion concentration is crucial for understanding the behavior of dissolved ionic compounds. In the dissociation of \( \mathrm{MgF}_2 \), for every 1 mole of \( \mathrm{Mg}^{2+} \) produced, 2 moles of \( \mathrm{F}^- \) are formed. If you know one ion's concentration, you can determine the other using the dissociation ratio. For instance, if the concentration of \( [\mathrm{F}^-] \) is \( 2.3 \times 10^{-3} \ \mathrm{M} \), we find \( [\mathrm{Mg}^{2+}] \) by dividing \( [\mathrm{F}^-] \) by 2 because there are twice as many \( \mathrm{F}^- \) ions.
  • \( [\mathrm{Mg}^{2+}] = \frac{1}{2} \times [\mathrm{F}^-] = 1.15 \times 10^{-3} \ \mathrm{M} \)
Understanding these concentrations is key when calculating other aspects of the solution, such as the \( K_{sp} \).

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Most popular questions from this chapter

Write the solubility product expression for each of the following slightly soluble ionic compounds in a saturated aqueous solution: (a) \(\mathrm{AgI}(s) \rightleftarrows \mathrm{Ag}^{+}(a q)+\mathrm{I}^{-}(a q)\) (b) \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}(\mathrm{~s}) \rightleftarrows 2 \mathrm{Ag}^{+}(a q)+\mathrm{CrO}_{4}^{2-}(a q)\) (c) \(\mathrm{Ag}_{3} \mathrm{PO}_{4}(s) \rightleftarrows 3 \mathrm{Ag}^{+}(a q)+\mathrm{PO}_{4}^{3-}(a q)\)

Write the general equilibrium constant expression for each of the following: (a) \(3 \mathrm{~A} \rightleftarrows 2 \mathrm{C}\) (b) \(A+B \rightleftarrows 2 C\) (c) \(3 \mathrm{~A}+5 \mathrm{~B} \rightleftarrows \mathrm{C}+4 \mathrm{D}\)

Explain why zinc carbonate, \(\mathrm{ZnCO}_{3}\), is more soluble in hydrochloric acid than in water.

Cadmium sulfide dissociates slightly in an aqueous solution as follows: $$ \mathrm{CdS}(s) \rightleftarrows \mathrm{Cd}^{2+}(a q)+\mathrm{S}^{2-}(a q) $$ Predict the direction of equilibrium shift for each of the following stresses: (a) increase \(\left[\mathrm{Cd}^{2+}\right]\) (b) increase \(\left[\mathrm{S}^{2-}\right]\) (c) decrease \(\left[\mathrm{Cd}^{2+}\right]\) (d) decrease \(\left[\mathrm{S}^{2-}\right]\) (e) add solid \(\mathrm{CdS}\) (f) add solid \(\mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2}\) (g) add solid \(\mathrm{NaNO}_{3}\) (h) add \(\mathrm{H}^{+}\)

The cobalt(II) ion concentration in a saturated solution of cobalt(II) sulfide, \(\operatorname{CoS},\) is \(7.7 \times 10^{-11} \mathrm{M}\). Calculate the value for the equilibrium constant.
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