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Magnesium fluoride dissolves in water to the extent of \(8.0 \times 10^{-2} \mathrm{~g} / \mathrm{L}\) at \(25 \mathrm{C}\). Calculate the solubility of \(\mathrm{MgF}_{2}(s)\) in moles per liter, and calculate \(K_{\mathrm{sp}}\) for \(\mathrm{MgF}_{2}\) at 25 .

Short Answer

Expert verified
The solubility of MgF鈧 in moles per liter is \(1.28 \times 10^{-3} \mathrm{~M}\) and the solubility product constant (K鈧涒倸) for MgF鈧 at 25 掳C is \(8.39 \times 10^{-9}\).

Step by step solution

01

Determine the molar mass of MgF鈧

To convert the solubility from g/L to moles per liter, we need to know the molar mass of MgF鈧. The molar masses of magnesium (Mg) and fluorine (F) are 24.31 g/mol and 19.00 g/mol, respectively. MgF鈧 contains one magnesium atom and two fluorine atoms, so we can calculate the molar mass as follows: Molar mass of MgF鈧 = (1 脳 24.31 g/mol) + (2 脳 19.00 g/mol) = 24.31 g/mol + 38.00 g/mol = 62.31 g/mol
02

Convert solubility from g/L to moles per liter

We are given the solubility of MgF鈧 as 8.0 脳 10鈦宦 g/L. Using the molar mass calculated in step 1, we can convert the solubility to moles per liter: Solubility (in moles per liter) = (8.0 脳 10鈦宦 g/L) / (62.31 g/mol) = 1.28 脳 10鈦宦 moles per liter
03

Write the balanced dissolution equation

Now, we will write the balanced equation for the dissolution of MgF鈧 in water: MgF鈧(s) 鈬 Mg虏鈦(aq) + 2F鈦(aq)
04

Set up an expression for K鈧涒倸

According to the balanced equation, the mole ratio of Mg虏鈦 to F鈦 is 1:2. So, for every mole of MgF鈧 that dissolves, one mole of Mg虏鈦 and two moles of F鈦 are produced in the solution. We can write the solubility in terms of the molar concentrations of the ions: [Mg虏鈦篯 = 1.28 脳 10鈦宦 M [F鈦籡 = 2 脳 1.28 脳 10鈦宦 M = 2.56 脳 10鈦宦 M The expression for the K鈧涒倸 is the product of the molar concentrations of the ions, raised to the power of their coefficients in the balanced equation: K鈧涒倸 = [Mg虏鈦篯 [F鈦籡虏
05

Calculate the K鈧涒倸 for MgF鈧

Now, we can substitute the molar concentrations determined in Step 4 into the K鈧涒倸 expression: K鈧涒倸 = (1.28 脳 10鈦宦) 脳 (2.56 脳 10鈦宦)虏 = (1.28 脳 10鈦宦) 脳 (6.5536 脳 10鈦烩伓) = 8.3886 脳 10鈦烩伖 So, the solubility product constant (K鈧涒倸) for MgF鈧 at 25 掳C is 8.39 脳 10鈦烩伖.

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Key Concepts

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

Solubility Calculation
Understanding solubility is fundamental in chemistry, especially when it comes to predicting whether a substance will dissolve in a given solvent and to what extent. Solubility calculation involves determining how much of a solute can be dissolved in a solvent to form a saturated solution at a defined temperature and pressure. In the context of the exercise provided, the solubility of magnesium fluoride (MgF2) in water was initially presented in grams per liter, which is a common unit for expressing solubility.

To work with solubility in chemical calculations, however, we often need to express it in moles per liter (M), which is a unit of concentration. This requires us to convert the mass of the solute to moles, a step that necessitates the knowledge of the molar mass of MgF2. Once we have the solute's solubility in moles per liter, we can engage in various calculations, including determining the solubility product constant (Ksp), which is a valuable expression of solubility in the realm of equilibrium chemistry.
Molar Mass
Molar mass is a fundamental concept in chemistry that represents the mass of one mole of a substance, usually expressed in grams per mole (g/mol). It's integral to converting between the mass of a substance and the number of moles, which is critical for stoichiometry and for understanding the composition of molecules and compounds. In our case, we calculated the molar mass of MgF2 by summing the molar masses of one magnesium atom and two fluorine atoms.

Determining Molar Mass:

For an element, the molar mass directly corresponds to the atomic weight found on the periodic table. For a compound like MgF2, you multiply the atomic weight of each element by the number of atoms of that element in the compound and then add the totals together. This process is essential for solubility calculations as it allows us to measure the amount of the substance in terms of its moles, which is essential for relating the mass of a solute to its corresponding number of particles.
Equilibrium Constant
The equilibrium constant is a crucial aspect of chemical equilibrium, expressing the ratio of product concentrations to reactant concentrations, with each raised to the power of their stoichiometric coefficients. The equilibrium constant provides insight into the extent of a reaction at equilibrium鈥攁nd in the case of soluble ionic compounds, we specifically use the solubility product constant (Ksp).

Ksp is unique because it applies to the special case of a solid dissolving into its constituent ions in a solution. When we calculated Ksp for MgF2, we focused on the ions produced when MgF2 dissolves. The small value of Ksp (in this case, 8.39 脳 10鈭9) indicates a low solubility, which is typical for sparingly soluble salts like MgF2. Knowledge of Ksp values enables chemists to predict whether a precipitation reaction will occur and to calculate concentrations of ions in a solution at equilibrium.

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Most popular questions from this chapter

Write the equilibrium expression for each of the following heterogeneous equilibria. a. \(4 \mathrm{Al}(s)+3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{Al}_{2} \mathrm{O}_{3}(s)\) b. \(\mathrm{NH}_{3}(g)+\mathrm{HCl}(g) \rightleftharpoons \mathrm{NH}_{4} \mathrm{Cl}(s)\) c. \(2 \mathrm{Mg}(s)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{MgO}(s)\)

Write the equilibrium expression for each of the following reactions. a. \(\mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g)\) b. \(2 \mathrm{NO}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g)\) c. \(\mathrm{P}_{4}(g)+6 \mathrm{Br}_{2}(g) \rightleftharpoons 4 \mathrm{PBr}_{3}(g)\)

According to Le Ch芒telier's principle, when a large excess of a gaseous reactant is added to a reaction system at equilibrium, the amounts of products

The equilibrium constant for the reaction $$ \mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) \rightleftharpoons 2 \mathrm{HF}(g) $$ has the value \(2.1 \times 10^{3}\) at a particular temperature. When the system is analyzed at equilibrium at this temperature, the concentrations of both \(\mathrm{H}_{2}(g)\) and \(\mathrm{F}_{2}(g)\) are found to be \(0.0021 M\). What is the concentration of \(\mathrm{HF}(g)\) in the equilibrium system under these conditions?

For the reaction $$ 2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) $$ \(K=2.4 \times 10^{-3}\) at a given temperature. At equilibrium it is found that \(\left[\mathrm{H}_{2} \mathrm{O}(g)\right]=1.1 \times 10^{-1} M\) and \(\left[\mathrm{H}_{2}(g)\right]=1.9 \times 10^{-2} M .\) What is the concentration of \(\mathrm{O}_{2}(g)\) under these conditions?

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