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(a) Explain the difference between solubility and solubility-product constant. (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) Solubility represents the maximum amount of a solute that can dissolve in a solvent (given in units like g/L or mol/L), while the solubility-product constant (\(K_{sp}\)) is an equilibrium constant representing the extent of an ionic compound's dissolution in water (unitless). (b) For the given ionic compounds, the expressions for the solubility-product constants are: \(K_{sp}(\mathrm{MnCO}_3) = [\mathrm{Mn^{2+}}] [\mathrm{CO_3^{2-}}]\) \(K_{sp}(\mathrm{Hg(OH)_2}) = [\mathrm{Hg^{2+}}] [\mathrm{OH^-}]^2\) \(K_{sp}(\mathrm{Cu_{3}(PO_4)_2}) = [\mathrm{Cu^{2+}}]^3 [\mathrm{PO_4^{3-}}]^2\)

Step by step solution

01

Definition of Solubility

Solubility is the maximum amount of a solute that can dissolve in a certain amount of solvent at a specific temperature and pressure. It is generally given in units such as grams of solute per liter of solvent (g/L) or moles of solute per liter of solvent (mol/L).
02

Definition of Solubility-Product Constant

The solubility-product constant, denoted by \(K_{sp}\), is an equilibrium constant that represents the extent to which an ionic compound dissolves in water to form a saturated solution. It is equal to the product of the concentrations of the ions formed, each raised to the power of their stoichiometric coefficients, at equilibrium.
03

Difference Between Solubility and Solubility-Product Constant

The main difference between solubility and solubility-product constant is that solubility represents the maximum amount of solute that can dissolve in a solvent, while the solubility-product constant represents the equilibrium constant of the dissolution process. Also, solubility has specific units (e.g., g/L or mol/L), while the solubility-product constant is unitless.
04

Dissolution Equation and \(K_{sp}\) Expression for \(\mathrm{MnCO}_3\)

The dissolution equation for \(\mathrm{MnCO}_3\) is: \(\mathrm{MnCO}_3 (s) \rightleftharpoons \mathrm{Mn^{2+}} (aq) + \mathrm{CO_3^{2-}} (aq)\) The expression for \(K_{sp}\) of \(\mathrm{MnCO}_{3}\) is: \(K_{sp} = [\mathrm{Mn^{2+}}] [\mathrm{CO_3^{2-}}]\)
05

Dissolution Equation and \(K_{sp}\) Expression for \(\mathrm{Hg}(\mathrm{OH})_{2}\)

The dissolution equation for \(\mathrm{Hg}(\mathrm{OH})_{2}\) is: \(\mathrm{Hg}(\mathrm{OH})_2 (s) \rightleftharpoons \mathrm{Hg^{2+}} (aq) + 2 \mathrm{OH^-} (aq)\) The expression for \(K_{sp}\) of \(\mathrm{Hg}(\mathrm{OH})_{2}\) is: \(K_{sp} = [\mathrm{Hg^{2+}}] [\mathrm{OH^-}]^2\)
06

Dissolution Equation and \(K_{sp}\) Expression for \(\mathrm{Cu}_{3}\left(\mathrm{PO}_{4}\right)_{2}\)

The dissolution equation for \(\mathrm{Cu}_{3}\left(\mathrm{PO}_4\right)_2\) is: \(\mathrm{Cu}_{3}\left(\mathrm{PO}_{4}\right)_{2} (s) \rightleftharpoons 3 \mathrm{Cu^{2+}} (aq) + 2 \mathrm{PO_4^{3-}} (aq)\) The expression for \(K_{sp}\) of \(\mathrm{Cu}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) 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
Imagine pouring sugar into a cup of tea; there is a limit to how much will dissolve, and that limit is what we call solubility. It represents how much solute can be dissolved in a solvent to form a homogeneous mixture at a specific temperature and pressure. Usually, solubility is expressed in terms like grams per liter (g/L) or moles per liter (mol/L). This concept is significant because it dictates the extent to which a substance can be dissolved in a solvent before reaching a saturated solution.

For instance, when we stir a spoonful of sugar in water, we're witnessing an increase in the solubility of sugar until no more can dissolve, reaching what is known as saturation. Beyond this point, any additional sugar will just settle at the bottom. With temperature changes, solubility can also vary; often, heat will increase solubility for many substances allowing more to dissolve.
Equilibrium Constant
To delve into the equilibrium constant, we must understand that chemical reactions can reach a state where the forward and reverse reactions occur at the same rate. When that happens, we've reached equilibrium. The equilibrium constant, abbreviated as 'K', gives us a peek into the balance between reactants and products in this steady state.

For the dissolution of ionic compounds in water, we focus on a special type of equilibrium constant known as the solubility-product constant, symbolized by (K_{sp}). It has no units and considers only the ions that are formed from the ionic compound in a saturated solution. Higher (K_{sp}) values signify greater solubility, and hence, a greater degree of dissociation into ions. This key piece of information is crucial in predicting the formation of a precipitate, understanding solution behavior, and manipulating separation processes like selective precipitation.
Dissolution Equations
Dissolution equations are the backbone of understanding how compounds separate into ions in a solution. Simply put, these equations provide a visual formula of the disintegration process of compounds when they dissolve in a solvent. Let's break down a complex process into an easy concept. Think of dropping a sugar cube into a cup of warm water. As it dissolves, the sugar molecules separate and spread throughout the water until evenly distributed.

In the case of ionic compounds, such as salt (NaCl), the dissolution process involves separation into ions (Na+ and Cl-). These ions will individually dissolve in water. The dissolution equation depicts this process, and when combined with the concept of the equilibrium constant, or (K_{sp}), we get a quantitative measure of this solubility behavior. Consequently, these dissolution equations and their constants are integral in predicting and explaining the outcomes of many reactions in aqueous solutions.

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Most popular questions from this chapter

A concentration of \(10-100\) parts per billion (by mass) of \(\mathrm{Ag}^{+}\) is an effective disinfectant in swimming pools. However, if the concentration exceeds this range, the \(\mathrm{Ag}^{+}\) can cause adverse health effects. One way to main-tain an appropriate concentration of \(\mathrm{Ag}^{+}\) is to add a slightly soluble salt to the pool. Using \(K_{s p}\) values from Appendix \(\mathrm{D}\), calculate the equilibrium concentration of \(\mathrm{Ag}^{+}\) in parts per billion that would exist in equilibrium with (a) \(\mathrm{AgCl}\), (b) \(\mathrm{AgBr}\), (c) AgI.

An unknown solid is entirely soluble in water. On addition of dilute \(\mathrm{HCl}\), a precipitate forms. After the precipitate is filtered off, the \(\mathrm{pH}\) is adjusted to about 1 and \(\mathrm{H}_{2} \mathrm{~S}\) is bubbled in; a precipitate again forms. After filtering off this precipitate, the \(\mathrm{pH}\) is adjusted to 8 and \(\mathrm{H}_{2} \mathrm{~S}\) is again added; no precipitate forms. No precipitate forms upon addition of \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{HPO}_{4} .\) The remaining solution shows a yellow color in a flame test. Based on these observations, which of the following compounds might be present, which are definitely present, and which are definitely absent: \(\mathrm{CdS}, \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}, \mathrm{HgO}, \mathrm{ZnSO}_{4}, \mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2}\) and \(\mathrm{Na}_{2} \mathrm{SO}_{4} ?\)

For each of the following slightly soluble salts, write the net ionic equation, if any, for reaction with acid: (a) MnS, (b) \(\mathrm{PbF}_{2}\), (c) \(\mathrm{AuCl}_{3}\) (d) \(\mathrm{Hg}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\), (e) \(\mathrm{CuBr}\).

A buffer is prepared by adding \(7.00 \mathrm{~g}\) of ammonia \(\left(\mathrm{NH}_{3}\right)\) and \(20.0 \mathrm{~g}\) of ammonium chloride \(\left(\mathrm{NH}_{4} \mathrm{Cl}\right)\) to enough water to form \(2.50 \mathrm{~L}\) of solution. (a) What is the \(\mathrm{pH}\) of this buffer? (b) Write the complete ionic equation for the reaction that occurs when a few drops of nitric acid are added to the buffer. (c) Write the complete ionic equation for the reaction that occurs when a few drops of potassium hydroxide solution are added to the buffer.

A 1.00-L solution saturated at \(25^{\circ} \mathrm{C}\) with lead(II) iodide contains \(0.54 \mathrm{~g}\) of \(\mathrm{PbI}_{2}\). Calculate the solubility- product constant for this salt at \(25^{\circ} \mathrm{C}\).

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