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(a) True or false: "solubility" and "solubility-product constant" are the same number for a given compound. (b) Write the expression for the solubility- product constant for each of the following ionic compounds: \(\mathrm{MnCO}_{3}, \mathrm{Hg}(\mathrm{OH})_{2},\) and \(\mathrm{Cu}_{3}\left(\mathrm{PO}_{4}\right)_{2}\).

Short Answer

Expert verified
(a) The statement is false, as solubility and solubility-product constant are not the same number for a given compound. (b) The solubility-product constant expressions are as follows: \(K_{sp}(\mathrm{MnCO}_{3}) = [\mathrm{Mn}^{2+}][\mathrm{CO}_{3}^{2-}]\), \(K_{sp}(\mathrm{Hg}(\mathrm{OH})_{2}) = [\mathrm{Hg}^{2+}][\mathrm{OH}^-]^2\), and \(K_{sp}(\mathrm{Cu}_{3}\left(\mathrm{PO}_{4}\right)_{2}) = [\mathrm{Cu}^{2+}]^3[\mathrm{PO}_{4}^{3-}]^2\).

Step by step solution

01

Part (a) - Solubility vs Solubility-Product Constant

Solubility is a measure of how much solute can dissolve in a given solvent, typically measured in moles per liter (M). The solubility of a compound depends on temperature, pressure, and the nature of the solvent. The solubility-product constant (Ksp) is a value that represents the equilibrium between a solid ionic compound and its dissolved ions. The Ksp of a compound is a constant value for a given temperature. Therefore, the solubility and solubility-product constant are not the same number for a given compound. So, the statement is false.
02

Part (b) - Expression for Solubility-Product Constant for MnCO3

The dissolution of MnCO3 in water can be represented as: \[\mathrm{MnCO}_{3}(s) \rightleftharpoons \mathrm{Mn}^{2+}(aq) + \mathrm{CO}_{3}^{2-}(aq)\] The solubility-product constant expression (Ksp) for MnCO3 is: \[K_{sp} = [\mathrm{Mn}^{2+}][\mathrm{CO}_{3}^{2-}]\]
03

Part (b) - Expression for Solubility-Product Constant for Hg(OH)2

The dissolution of Hg(OH)2 in water can be represented as: \[\mathrm{Hg}(\mathrm{OH})_{2}(s) \rightleftharpoons \mathrm{Hg}^{2+}(aq) + 2\mathrm{OH}^{-}(aq)\] The solubility-product constant expression (Ksp) for Hg(OH)2 is: \[K_{sp} = [\mathrm{Hg}^{2+}][\mathrm{OH}^-]^2\]
04

Part (b) - Expression for Solubility-Product Constant for Cu3(PO4)2

The dissolution of Cu3(PO4)2 in water can be represented as: \[\mathrm{Cu}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s) \rightleftharpoons 3\mathrm{Cu}^{2+}(aq) + 2\mathrm{PO}_{4}^{3-}(aq)\] The solubility-product constant expression (Ksp) for Cu3(PO4)2 is: \[K_{sp} = [\mathrm{Cu}^{2+}]^3[\mathrm{PO}_{4}^{3-}]^2\]

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Key Concepts

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

Solubility
Solubility describes how much of a substance can be dissolved in a solvent at a specific temperature and pressure. For example, if we say that the solubility of sugar in water is 200 grams per liter at room temperature, it means you can dissolve 200 grams of sugar in one liter of water under those conditions. When discussing solubility, remember it can change with temperature or when different solvents are used.
Sometimes, increasing the temperature can increase the solubility of solids in liquids, which is commonly seen when making a hot cup of tea as more sugar dissolves when the water is hot.
However, solubility and the solubility-product constant (Ksp) are related but not the same, as solubility refers to an amount, while Ksp is a constant value used in equilibrium calculations.
Chemical equilibrium
Chemical equilibrium happens when a chemical reaction reaches a state where the concentrations of reactants and products remain constant over time. This does not mean the reaction has stopped; rather, it means the forward and reverse reactions occur at the same rate. For a solution in equilibrium, you can picture it as a balanced scale where no matter how you move it, the balance between reactants and products remains unchanged.
In the equation \[ aA + bB \rightleftharpoons cC + dD \]the equilibrium expression is given by:\[ K = \frac{{[C]^c[D]^d}}{{[A]^a[B]^b}} \]where \[ [C], [D], [A], [B] \]are the concentrations of each species in the reaction mixture. At equilibrium, these concentrations do not change.
Ionic compounds
Ionic compounds are substances composed of positive and negative ions held together by strong electrostatic forces known as ionic bonds. These compounds often form crystalline solids and dissolve in water to produce free ions, enabling conduction of electricity in solution.
For example, common table salt, sodium chloride (NaCl), is an ionic compound. When dissolved in water, it separates into sodium (Na+) and chloride (Cl-) ions.
It's important to note that not all ionic compounds are highly soluble. Some can form precipitates, which are solids that do not dissolve well in water. Understanding how ionic compounds dissolve helps us write expressions for the solubility-product constant, which considers the concentration of ions in a solution at equilibrium.
Ksp expressions
The solubility-product constant (Ksp) is a special type of equilibrium constant useful in describing the solubility of ionic compounds.It is specifically used for those that dissolve to a very small extent. When an ionic compound dissolves in water, it dissociates into its ions.The Ksp expression represents this process in terms of the concentrations of the ions at equilibrium.
For instance, consider \( ext{MnCO}_3 \):
The dissolution is represented as \[ \text{MnCO}_3(s) \rightleftharpoons \text{Mn}^{2+}(aq) + \text{CO}_3^{2-}(aq) \]The expression for Ksp is \[ K_{sp} = [\text{Mn}^{2+}][\text{CO}_3^{2-}] \]The Ksp tells us the extent to which the compound will dissolve. A smaller value of Ksp indicates a less soluble compound.
Using Ksp helps chemists predict whether a precipitate will form in a given reaction and understand the conditions necessary to achieve equilibrium.

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Most popular questions from this chapter

For each statement, indicate whether it is true or false. (a) The solubility of a slightly soluble salt can be expressed in units of moles per liter. (b) The solubility product of a slightly soluble salt is simply the square of the solubility. (c) The solubility of a slightly soluble salt is independent of the presence of a common ion. (d) The solubility product of a slightly soluble salt is independent of the presence of a common ion.

Write the expression for the solubility-product constant for each of the following ionic compounds: \(\mathrm{BaCrO}_{4}, \mathrm{CuS}, \mathrm{PbCl}_{2}\) and \(\mathrm{LaF}_{3} .\)

Suggest how the cations in each of the following solution mixtures can be separated: (a) \(\mathrm{Na}^{+}\) and \(\mathrm{Cd}^{2+},(\mathbf{b}) \mathrm{Cu}^{2+}\) and \(\mathrm{Mg}^{2+},(\mathbf{c}) \mathrm{Pb}^{2+}\) and \(\mathrm{Al}^{3+},(\mathbf{d}) \mathrm{Ag}^{+}\) and \(\mathrm{Hg}^{2+} .\)

Which of these statements about the common-ion effect is most correct? (a) The solubility of a salt MA is decreased in a solution that already contains either \(\mathrm{M}^{+}\) or \(\mathrm{A}^{-} .\) (b) Common ions alter the equilibrium constant for the reaction of an ionic solid with water. \((\mathbf{c})\) The common-ion effect does not apply to unusual ions like \(\mathrm{SO}_{3}^{2-} .(\mathbf{d})\) The solubility of a salt MA is affected equally by the addition of either \(\mathrm{A}^{-}\) or a noncommon ion.

How many microliters of \(1.000 \mathrm{M} \mathrm{NaOH}\) solution must be added to \(25.00 \mathrm{~mL}\) of a \(0.1000 \mathrm{M}\) solution of lactic acid \(\left[\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{COOH}\right.\) or \(\left.\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{3}\right]\) to produce a buffer with \(\mathrm{pH}=3.75 ?\)

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