![]() ![]() We can establish this equilibrium either by adding solid calcium carbonate to water or by mixing a solution that contains calcium ions with a solution that contains carbonate ions. However, when we add an excess of solid AgCl to water, it dissolves to a small extent and produces a mixture consisting of a very dilute solution of Ag + and Cl – ions in equilibrium with undissolved silver chloride: Recall from the solubility rules in an earlier chapter that halides of Ag + are not normally soluble. Silver chloride is what’s known as a sparingly soluble ionic solid (Figure 17.6a). We will also learn how to use the equilibrium constant of the reaction to determine the concentration of ions present in a solution. In this section, we will find out how we can control the dissolution of a slightly soluble ionic solid by the application of Le Châtelier’s principle. We want the calcium carbonate in a chewable antacid to dissolve because the CO 3 2- ions produced in this process help soothe an upset stomach. On the other hand, sometimes we want a substance to dissolve. Preventing the dissolution prevents the decay. ![]() The dissolution process is aided when bacteria in our mouths feast on the sugars in our diets to produce lactic acid, which reacts with the hydroxide ions in the calcium hydroxylapatite. Tooth decay, for example, occurs when the calcium hydroxylapatite, which has the formula Ca 5(PO 4) 3(OH), in our teeth dissolves. In some cases, we want to prevent dissolution from occurring. The preservation of medical laboratory blood samples, mining of sea water for magnesium, formulation of over-the-counter medicines such as Milk of Magnesia and antacids, and treating the presence of hard water in your home’s water supply are just a few of the many tasks that involve controlling the equilibrium between a slightly soluble ionic solid and an aqueous solution of its ions. Carry out equilibrium computations involving solubility, equilibrium expressions, and solute concentrations.Write chemical equations and equilibrium expressions representing solubility equilibria.The answer is rounded to one significant figure, the number of sig figs you have for the volume of water. that sodium bromide has a molar solubility of #7.07 * 10^(-4)#"mol L"^(-1)#.Ĭonsequently, you can say that #"100 L"# of water, which you can easily approximate to be equal to the volume of the solution, of a saturated silver bromide solution, will hold This tells you that you can only hope to dissolve #7.07 * 10^(-4)# moles of silver bromide for every #"1 L"# of solution at the given temperature, i.e. Now, if you take #s# to represent the concentration of silver bromide that dissociates to produce silver cations and bromide anions in the saturated solution, you can say that you have #"AgBr"_ ((s)) rightleftharpoons "Ag"_ ((aq))^(+) + "Br"_ ((aq))^(-)#īy definition, the solubility product constant, #K_(sp)#, for silver bromide is equal to So, you know that silver bromide is considered insoluble in water, so you can say that when this salt is dissolved in water, an equilibrium is established between the undissolved solid and the solvated ions. the number of moles of silver bromide that can dissolve in #"1 L"# of solution to produce a saturated solution. The first thing that you need to do here is to figure out the molar solubility of silver bromide in water, i.e. ![]()
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