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Chapter 16: Problem 10

Solubility is an equilibrium position, whereas \(K_{\mathrm{sp}}\) is an equilibrium constant. Explain the difference.

Short Answer

Expert verified
Solubility refers to the maximum amount of solute that can dissolve in a given amount of solvent at a specific temperature to form a saturated solution, whereas the solubility product constant (Kₛₚ) represents the equilibrium position for the dissociation of a slightly soluble ionic compound into its respective ions in a saturated solution. Solubility is generally expressed in terms of concentration, while Kₛₚ is a constant that indicates how "dissolved" the solute is in the solution, signifying the extent of dissociation of the solute at a specific temperature.

Step by step solution

01

Define Solubility

Solubility is a term that refers to the maximum amount of solute that can dissolve in a given amount of solvent at a specific temperature, to form a saturated solution. In other words, solubility defines the maximum concentration of the solute in the solution.
02

Define Solubility Product Constant (Kₛₚ)

The solubility product constant (Kₛₚ) is an equilibrium constant that represents the extent to which a slightly soluble ionic compound dissociates into its respective ions in a saturated solution. The constant depends on the temperature and ionic strength of the solution.
03

Understand the Dissociation of the Solid Solute

In a saturated solution, the solute dissociates into its corresponding ions according to the following general equation: \(AB_{(s)} \rightleftharpoons A^{+}_{(aq)} + B^{-}_{(aq)}\) Here, AB represents the solid solute, and A⁺ and B⁻ are the cations and anions in the solution.
04

Equilibrium Constant for the Dissociation

The equilibrium constant for the dissociation of the slightly soluble solute is given by: \[K_{\mathrm{sp}} = [A^{+}][B^{-}]\] where Kₛₚ is the solubility product constant, and [A⁺] and [B⁻] are the molar concentrations of the respective ions at equilibrium.
05

Compare Solubility and Kₛₚ

Solubility tells us the maximum concentration of the solute in the saturated solution at a particular temperature. It is usually expressed in terms like grams of solute per 100 grams of solvent or as molar concentration (mol/L). On the other hand, Kₛₚ is a constant related to the equilibrium position of the solute's dissociation and expresses how "dissolved" the solute is in the solution. Kₛₚ gives an idea about the extent to which the dissociation of AB(s) occurs in the solution at a specific temperature.

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Most popular questions from this chapter

When \(\mathrm{Na}_{3} \mathrm{PO}_{4}(a q)\) is added to a solution containing a metal ion and a precipitate forms, the precipitate generally could be one of two possibilities. What are the two possibilities?

Which of the following will affect the total amount of solute that can dissolve in a given amount of solvent? a. The solution is stirred. b. The solute is ground to fine particles before dissolving. c. The temperature changes.

Write balanced equations for the dissolution reactions and the corresponding solubility product expressions for each of the following solids. a. \(\mathrm{AgC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) b. \(\mathrm{Al}(\mathrm{OH})_{3}\) c. \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\)

Calculate the molar solubility of \(\mathrm{Mg}(\mathrm{OH})_{2}, K_{\mathrm{sp}}=8.9 \times 10^{-12}\)

Calculate the equilibrium concentrations of \(\mathrm{NH}_{3}, \mathrm{Cu}^{2+}\), \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)^{2+}, \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{2}^{2+}, \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{3}^{2+}\), and \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}{ }^{2+}\) in a solution prepared by mixing \(500.0 \mathrm{~mL}\) of \(3.00 \mathrm{M} \mathrm{NH}_{3}\) with \(500.0 \mathrm{~mL}\) of \(2.00 \times 10^{-3} M \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\). The stepwise equilib- ria are \(\mathrm{Cu}^{2+}(a q)+\mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{CuNH}_{3}^{2+}(a q)\) \(K_{1}=1.86 \times 10^{4}\) \(\mathrm{CuNH}_{3}{ }^{2+}(a q)+\mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{2}^{2+}(a q)\) \(K_{2}=3.88 \times 10^{3}\) \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{2}{ }^{2+}(a q)+\mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{3}^{2+}(a q)\) \(K_{3}=1.00 \times 10^{3}\) \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{3}{ }^{2+}(a q)+\mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}{ }^{2+}(a q)\) \(K_{4}=1.55 \times 10^{2}\)
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