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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 (K鈧涒倸) 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 K鈧涒倸 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 (K鈧涒倸)

The solubility product constant (K鈧涒倸) 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鈦 and B鈦 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 K鈧涒倸 is the solubility product constant, and [A鈦篯 and [B鈦籡 are the molar concentrations of the respective ions at equilibrium.
05

Compare Solubility and K鈧涒倸

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, K鈧涒倸 is a constant related to the equilibrium position of the solute's dissociation and expresses how "dissolved" the solute is in the solution. K鈧涒倸 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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Key Concepts

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

Equilibrium Constant
In chemistry, the term 'equilibrium constant' is fundamental to understanding how chemical reactions progress to a state of balance. In a reversible reaction, where reactants convert to products and vice versa, equilibrium is achieved when the rates of the forward and backward reactions are equal. At this point, the concentration of reactants and products remains constant over time. The equilibrium constant, denoted as \(K\), provides a quantitative measure of the relative concentrations of these reactants and products at equilibrium. This constant is a characteristic property of the particular chemical reaction at a given temperature.

The value of \(K\) sheds light on the favorability of a reaction. A large \(K\) value indicates that the products are favored at equilibrium, meaning there is a higher concentration of products compared to reactants. Conversely, a small \(K\) suggests that the reactants are predominant. The equilibrium constant is derived from the concentrations of the products and reactants, each raised to the power of their stoichiometric coefficients in the balanced chemical equation:

\[K = \frac{[products]}{[reactants]}\]Remember, equilibrium constants are temperature-specific, so changes in temperature can alter the value of \(K\). Understanding the equilibrium constant helps in predicting the position of equilibrium and the extent of a reaction under specific conditions.
Solubility Product Constant
The solubility product constant, often abbreviated as \(K_{sp}\), is a special type of equilibrium constant used for sparingly soluble ionic compounds. It reflects the degree to which a compound can dissociate into its ions when in a saturated solution. Essentially, \(K_{sp}\) gives us insight into the solubility of a compound.

For a general reaction where a solid ionic compound, such as \(AB\), dissolves to produce its ions:

\[AB_{(s)} \rightleftharpoons A^{+}_{(aq)} + B^{-}_{(aq)}\]The solubility product constant is expressed as:

\[K_{sp} = [A^{+}][B^{-}]\]Here, \([A^{+}]\) and \([B^{-}]\) represent the molar concentrations of the ions at equilibrium. The value of \(K_{sp}\) shows how many ions can exist in the solution before the compound starts to precipitate. A greater \(K_{sp}\) value indicates higher solubility, while a smaller \(K_{sp}\) suggests that the compound is less likely to dissolve.

Changes in temperature can affect \(K_{sp}\) just like other equilibrium constants, usually increasing solubility with a rise in temperature. The knowledge of \(K_{sp}\) is crucial in fields like predicting and controlling the precipitate formation in chemical processes.
Dissociation of Ionic Compounds
The dissociation of ionic compounds is a key concept in understanding how these substances behave in solutions. An ionic compound like \(AB\) dissociates into its constituent ions when it dissolves in water. This process involves overcoming the ionic bonds that hold the compound together in its solid form, allowing the anions and cations to disperse into the solvent.

When dissolved, the compound separates according to the equation:

\[AB_{(s)} \rightleftharpoons A^{+}_{(aq)} + B^{-}_{(aq)}\]This equation shows the dynamic equilibrium between the solid phase and the solvated ions. In a saturated solution, the rate at which \(AB\) dissolves to form \(A^+\) and \(B^-\) is equal to the rate at which \(A^+\) and \(B^-\) recombine to form \(AB\). At equilibrium, the concentrations of \(A^+\) and \(B^-\) remain constant.

Understanding dissociation is vital for predicting the behavior of ionic compounds in various solutions. It also aids in calculating important parameters like molar concentration and conducting titration experiments. For instance, the complete dissociation of acidic or basic compounds in water determines their strength and reactivity in chemical reactions. Recognizing how ionic compounds dissociate helps in fields ranging from environmental science to chemical engineering.

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Most popular questions from this chapter

Calculate the molar solubility of \(\mathrm{Co}(\mathrm{OH})_{3}, K_{\mathrm{sp}}=2.5 \times 10^{-43}\)

Calculate the solubility of each of the following compounds in moles per liter. Ignore any acid-base properties. a. \(\mathrm{PbI}_{2}, K_{\mathrm{sp}}=1.4 \times 10^{-8}\) b. \(\mathrm{CdCO}_{3}, K_{\text {sp }}=5.2 \times 10^{-12}\) c. \(\mathrm{Sr}_{3}\left(\mathrm{PO}_{4}\right)_{2}, K_{\text {sp }}=1 \times 10^{-31}\)

The active ingredient of Pepto-Bismol is the compound bismuth subsalicylate, which undergoes the following dissociation when added to water: \(\begin{aligned} \mathrm{C}_{7} \mathrm{H}_{5} \mathrm{BiO}_{4}(s)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{C}_{7} \mathrm{H}_{4} \mathrm{O}_{3}{ }^{2-}(a q) \\\ &+\mathrm{Bi}^{3+}(a q)+\mathrm{OH}^{-}(a q) \quad K=? \end{aligned}\) 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.

For each of the following pairs of solids, determine which solid has the smallest molar solubility. a. \(\mathrm{FeC}_{2} \mathrm{O}_{4}, K_{\mathrm{sp}}=2.1 \times 10^{-7}\), or \(\mathrm{Cu}\left(\mathrm{IO}_{4}\right)_{2}, K_{<\rho}=1.4 \times 10^{-7}\) b. \(\mathrm{Ag}_{2} \mathrm{CO}_{3}, K_{\text {s }}=8.1 \times 10^{-12}\), or \(\mathrm{Mn}(\mathrm{OH})_{2}\), \(K_{\text {sp }}=2 \times 10^{-13}\)

Assuming that the solubility of \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)\) is \(1.6 \times 10^{-7}\) \(\mathrm{mol} / \mathrm{L}\) at \(25^{\circ} \mathrm{C}\), calculate the \(K_{\text {op }}\) for this salt. Ignore any potential reactions of the ions with water.

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