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Chapter 16: Problem 12

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

Short Answer

Expert verified
Solubility is the equilibrium position that describes the balance between dissolving and precipitation processes in a saturated solution and can be affected by factors such as temperature, pressure, and presence of other ions. It is expressed in units of concentration (e.g., g/L or mol/L). On the other hand, Ksp, the solubility product constant, is a dimensionless number that measures the extent of dissolution for a specific ionic compound at a given temperature and is constant regardless of the ion concentrations in the solution. Solubility and Ksp are related but describe different aspects of the dissolution process.

Step by step solution

01

Define Solubility

Solubility refers to the maximum amount of a solute that can dissolve in a solvent to form a saturated solution at a given temperature and pressure. It is an equilibrium position, which means it depends on the balance between the dissolving and precipitation processes of the solute in the solvent. Solubility can be expressed in various units, such as grams per liter (g/L) or moles per liter (mol/L).
02

Define Solubility Product Constant (Ksp)

The solubility product constant, often denoted as Ksp, is a measure of the extent of dissolution of a sparingly soluble ionic compound in water at a particular temperature. Ksp represents the product of the concentrations of the ions present in the saturated solution, each raised to the power of their respective stoichiometric coefficient. It is a constant value for a specific compound at a given temperature and does not change with the concentration of the ions in the solution.
03

Discuss the Differences

The main differences between solubility and Ksp are: 1. Solubility is an equilibrium position, while Ksp is an equilibrium constant. This means that solubility describes the balance between dissolving and precipitation processes, whereas Ksp quantifies the extent of dissolution for a particular substance at a specific temperature. 2. Solubility can be affected by various factors, such as temperature, pressure, and the presence of other ions in the solution, whereas Ksp is a constant value, specific to a particular ionic compound and temperature. 3. Solubility is often represented in units of concentration (e.g., g/L or mol/L), whereas Ksp is represented as a dimensionless number, since it is a product of concentrations raised to their stoichiometric coefficients. 4. Solubility and Ksp are related but not the same. Using Ksp and the stoichiometry of the dissolution reaction, one can calculate the solubility of a compound, but they are different concepts describing different aspects of the dissolution process.

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Most popular questions from this chapter

A mixture contains \(1.0 \times 10^{-3} M \mathrm{Cu}^{2+}\) and \(1.0 \times 10^{-3} M\) \(\mathrm{Mn}^{2+}\) and is saturated with 0.10\(M \mathrm{H}_{2} \mathrm{S} .\) Determine a pH where CuS precipitates but MnS does not precipitate. \(K_{\mathrm{sp}}\) for \(\mathrm{CuS}=8.5 \times 10^{-45}\) and \(K_{\mathrm{sp}} \mathrm{for} \mathrm{MnS}=2.3 \times 10^{-13} .\)

Calculate the final concentrations of \(\mathrm{K}^{+}(a q), \mathrm{C}_{2} \mathrm{O}_{4}^{2-}(a q),\) \(\mathrm{Ba}^{2+}(a q),\) and \(\operatorname{Br}^{-}(a q)\) in a solution prepared by adding 0.100 \(\mathrm{L}\) of \(0.200M\) \(\mathrm{K}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) to 0.150 \(\mathrm{L}\) of \(0.250 M\) \(\mathrm{BaBr}_{2}\) . (For \(\mathrm{BaC}_{2} \mathrm{O}_{4}, K_{\mathrm{sp}}=2.3 \times 10^{-8} . )\)

The active ingredient of Pepto-Bismol is the compound bismuth subsalicylate, which undergoes the following dissociation when added to water: $$\mathrm{C}_{7} \mathrm{H}_{5} \mathrm{BiO}_{4}(s)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{C}_{7} \mathrm{H}_{4} \mathrm{O}_{3}^{2-}(a q) +\mathrm{Bi}^{3+}(a q)+\mathrm{OH}^{-}(a q) \qquad K=?$$ If the maximum amount of bismuth subsalicylate that reacts by this reaction is \(3.2 \times 10^{-19} \mathrm{mol} / \mathrm{L}\) , calculate the equilibrium constant for the preceding reaction.

Calculate the solubility of \(\mathrm{Co}(\mathrm{OH})_{2}(s)\left(K_{\mathrm{sp}}=2.5 \times 10^{-16}\right)\) in a buffered solution with a \(\mathrm{pH}\) of 11.00.

In the presence of \(\mathrm{CN}^{-}, \mathrm{Fe}^{3+}\) forms the complex ion \(\mathrm{Fe}(\mathrm{CN})_{6}^{3-} .\) The equilibrium concentrations of \(\mathrm{Fe}^{3+}\) and \(\mathrm{Fe}(\mathrm{CN})_{6}^{3-}\) are \(8.5 \times 10^{-40} \mathrm{M}\) and \(1.5 \times 10^{-3} M,\) respectively, in a \(0.11-M\) KCN solution. Calculate the value for the overall formation constant of \(\mathrm{Fe}(\mathrm{CN})_{6}^{3-}\) . $$\mathrm{Fe}^{3+}(a q)+6 \mathrm{CN}^{-}(a q) \rightleftharpoons \mathrm{Fe}(\mathrm{CN})_{6}^{3-}(a q) \qquad K_{\text { overall }}=?$$
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