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A saturated solution of \(\mathrm{MgF}_{2}\) has \(\left[\mathrm{Mg}^{2+}\right]=2.6 \times 10^{-4} \mathrm{M}\) and \(\left[\mathrm{F}^{-}\right]=5.2 \times 10^{-4} \mathrm{M}\). What is the value of \(K_{\mathrm{sp}}\) for \(\mathrm{MgF}_{2}\) ?

Short Answer

Expert verified
The value of Ksp for MgF2 is approximately 7.03 x 10^{-11}.

Step by step solution

01

Write Down the Dissociation Equation

The first step in calculating the solubility product constant, or Ksp, for a salt is to write the balanced chemical dissociation equation. For magnesium fluoride (\text{MgF}_{2}), the dissociation in water is represented as:\[ \text{MgF}_{2}(s) \rightleftharpoons \text{Mg}^{2+}(aq) + 2\text{F}^{-}(aq) \]
02

Write the Expression for Ksp

The solubility product constant expression for this reaction is given by the formula:\[ K_{\text{sp}} = [\text{Mg}^{2+}][\text{F}^{-}]^{2} \]Here, the concentrations of the ions raised to the power of their coefficients in the balanced equation.
03

Substitute the Concentration Values into Ksp Expression

Plug in the given concentration values into the Ksp expression:\[ K_{\text{sp}} = (2.6 \times 10^{-4}) \times (5.2 \times 10^{-4})^{2} \]
04

Calculate Ksp

Perform the multiplication to find the Ksp value:\[ K_{\text{sp}} = (2.6 \times 10^{-4}) \times (5.2 \times 10^{-4})^{2} = (2.6 \times 10^{-4}) \times (2.704 \times 10^{-7}) \approx 7.03 \times 10^{-11} \]

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Key Concepts

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

Chemical Equilibrium
The concept of chemical equilibrium is pivotal in understanding how chemicals react and form new substances. It refers to a state in which the rate of the forward reaction equals the rate of the reverse reaction, thus leading to a balance in the concentrations of reactants and products over time. Imagine it as a busy intersection where cars—the molecules—flow in and out at the same time, maintaining a steady traffic flow.

In the context of solubility, a saturated solution represents a state of equilibrium between the dissolved ions and the undissolved solid. If you add more solid to the saturated solution, it won't dissolve because the system is at equilibrium; similarly, no more solid will precipitate unless you change the conditions of the solution.
Ksp Calculation
The process of Ksp calculation is based on the knowledge of chemical equilibrium. The solubility product constant (Ksp) is a special type of equilibrium constant that measures the solubility of an ionic compound in a solvent. The calculation is straightforward: it involves writing the dissociation equation, forming the Ksp expression, and then plugging in the molar concentrations of the ions.

An important point to note is that solid concentrations are not included in Ksp expressions because their activity is considered constant. Thus, Ksp exclusively involves the concentrations of dissolved ions. The calculated Ksp can give you valuable insight into the solubility of the compound—a smaller Ksp value corresponds to a less soluble substance.
Ionic Compounds Solubility
Ionic compounds solubility can vary greatly and is influenced by temperature and the presence of other ions in solution. The solubility product constant, Ksp, is an essential indicator that allows us to predict the solubility. A higher Ksp value suggests that the compound is more soluble, while a lower Ksp indicates that the substance is less soluble.

Furthermore, another concept to keep in mind is the common ion effect, which describes how the solubility of an ionic compound decreases when a common ion is added to the solution. For example, adding extra fluoride ions to a solution of magnesium fluoride will suppress the dissolution of MgF2 because the product of the ionic concentrations will reach the Ksp value more rapidly.
Saturated Solution Chemistry
In the realm of saturated solution chemistry, a solution is said to be 'saturated' when it contains the maximum concentration of a solute that can dissolve at a given temperature. In this equilibrium state, the rate of dissolution of the solute equals the rate of precipitation. Think of it as a dance floor at maximum capacity where for every person that leaves, another instantly takes their place.

For students dealing with problems related to saturated solutions, it's critical to recognize that changing the conditions of the solution—such as temperature or the addition of other substances—can 'shift' the equilibrium. This can lead to either more solute dissolving or precipitating out of the solution, hence altering the concentration of the ions in the solution and affecting the Ksp value.

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Most popular questions from this chapter

Consider the reaction. $$ \mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g) $$ An equilibrium mixture of this reaction at a certain temperature has \([\mathrm{CO}]=0.105 \mathrm{M},\left[\mathrm{H}_{2}\right]=0.114 \mathrm{M}\), and \(\left[\mathrm{CH}_{3} \mathrm{OH}\right]=0.185 \mathrm{M}\). What is the value of the equilibrium constant at this temperature?

A saturated solution of \(\mathrm{AgI}\) has \(\left[\mathrm{Ag}^{+}\right]=9.2 \times 10^{-9} \mathrm{M}\) and \(\left[\mathrm{I}^{-}\right]=9.2 \times 10^{-9} \mathrm{M}\). What is the value of \(K_{\mathrm{sp}}\) for \(\mathrm{AgI}\) ?

Consider the effect of a volume change on this reaction at equilibrium. $$ \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g) $$ Predict the effect (shift right, shift left, or no effect) of these changes. (a) increasing the reaction volume (b) decreasing the reaction volume

Consider the reaction. $$ \mathrm{Fe}^{3+}(a q)+\mathrm{SCN}^{-}(a q) \rightleftharpoons \mathrm{FeSCN}^{2+}(a q) $$ Asolution is made containing initial \(\left[\mathrm{Fe}^{3+}\right]=1.0 \times 10^{-3} \mathrm{M}\) and initial \(\left[\mathrm{SCN}^{-}\right]=8.0 \times 10^{-4} \mathrm{M}\). At equilibrium, \(\left[\mathrm{FeSCN}^{2+}\right]=1.7 \times 10^{-4} \mathrm{M}\). Calculate the value of the equilibrium constant. Hint: Use the chemical reaction stoichiometry to calculate the equilibrium concentrations of \(\mathrm{Fe}^{3+}\) and \(\mathrm{SCN}^{-}\).

Describe three ways a reaction at equilibrium can be changed such that it is no longer in equilibrium. For each case, indicate which way the reaction will shift to return to equilibrium.

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