RI 2023 Solubility Equilibria

Raffles Institution Year 6 H2 Chemistry 2023 Lecture Notes 20 - Solubility Equilibria A Contents (a) Solubility product (b) Common ion effect (c) Complex ion formation C Lecture Outline 1 . Solubility 2. Solubility product 3. Ionic product and precipitation 4. Common ion effect 5. Formation of complex ions 6. Effect of pH 7. Limitations to solubility product concept 1 I solubility 1.1 Defining solubility B Learning Outcomes At the end of the lectures , you should be able to: (a) show understanding of, and apply, the concept of solubility product, Ksp (b) calculate l<sp from concentrations and vice versa (c) discuss the effects on the solubility of ionic salts by the following: • common ion effect • formation of complex ions, as exemplified by the reactions of halide ions with aqueous silver ions followed by aqueous ammonia D References 1. Chemistry in Context by Hill & Holman 2. A level Chemistry by Ramsden 3. Chemistry : The Molecular Nature of Matter and Change by Martin S. Silberberg • The maximum amount of solute which can dissolve in a given amount of solvent at a particular temperature is called the solubility of the solute in that solvent. • A solution containing the maximum amount of solute that can be dissolved in the given amount of solvent is called a saturated solution. • The solubility of a salt at a stated temperature can be expressed in various units, e.g. Solubility may be expressed as: Units number of moles of solute dissolved in 1 dm3 of solution mol dm-3 mass of solute dissolved in 1 dm3 of solution g dm-3 mass of solute dissolved in 100 g of solvent g per 100 g solvent mass of solute dissolved in 106 g of solution ppm (parts per million) WorkeJ{ e~-~mpJicJi If the solubility of Ag2CrO4 is 6.05 x 10-5 mol dm-3, what is the concentration of each of its constituent ions in a saturated solution? Ag2CrQ4(s) .= 2Ag+(aq) + Cro/-(aq) In a saturated solution of Ag2CrQ4, [Ag+)= (2)(6.05 x 10-5) = 1.21 x 10-4 mol dm-3 [Cro/-J = 6.05 x 10-5 mol dm-3 -1-

1.2 Dissolution as an Equilibrium Process Soluble salts Soluble salts dissociate fully in solution into their constituent ions. Example: NaC/(s)~ Na•(aq) + C,(aq) Sparingly soluble salts • When a small amount of a sparingly soluble ionic solid, MX, dissolves in water at a given temperature, an aqueous solution containing M• and x- ions is formed. • As more solid MX is added to the solution, the concentrations of the ions increase. • Eventually, the solution becomes saturated, i.e. it contains the maximum amount of dissolved solute at that particular temperature in the presence of undissolved solute. • At this point, The ions in the saturated solution are in dvnamic equilibrium with the excess undissolved solid: ~-----------------------------, , In a saturated solution, : : dissolution and precipitation : : occur at the same rate. , ~------------------------------· The rate of the forward reaction equals rate of the backward reaction, and there is no net change in concentration of the ions. 2 I Solubility Product 2.1 Defining the Solubility Product • The equilibrium established in a saturated solution of a sparingly soluble salt with the general formula MaXt, is: • By applying the equilibrium law, the equilibrium constant, Kc, is given by • The concentration of a pure solid, which is proportional to its density, is constant at a particular temperature. Since [MaXb(s)] is constant at a given temperature, • Hence, we may define a new equilibrium constant called solubility product, Ksp . units: (mol dm-3)3•b • The solubility product, Ksp, for the sparingly soluble salt MaXb is the equilibrium constant for the equilibrium established between the undissolved salt and its constituent ions in a saturated solution. • Like other values of equilibrium constants, the value of Ksp varies only with temperature. -2-

A • The solubility products of some common compounds at 25 °C are shown below. Compound Solubility Product Compound Solubility Product BaSQ4 1.1 x 10-10 mol2 dm-6 AgC/ 1.8 x 10-10 mol2 dm-6 BaC03 2.6 x 10-9 mol2 dm-6 AgBr 5.4 x 10-13 mol2 dm-6 Caso. 5.0 x 10-5 mol2 dm-6 Agl 8.5 x 10-17 mol2 dm-6 CaCOJ 3.4 x 10-9 mol2 dm-6 NiS 4.0 x 10-21 mol2 dm-6 PbBr2 6.6 x 10-6 mol3 dm-9 ZnS 1.6 x 10-2• mol2 dm-6 PbC/2 1. 7 x 10-5 mol3 dm-9 PbS 1.3 X 10-25 mol2 dm-6 Pbh 9.8 x 10-9 mol3 dm-9 CuS 6.3 x 10-35 mol2 dm-6 Complete the table below. Sparingly Equation for the solubility soluble salt e uilibrium Expression for Ksp BaSQ4(S) 2.2 Relationship between solubility and solubility product • The solubility and solubility product of some compounds are shown below: Compound AgC/ AgBr Agl Ksp 1.8 X 1Q-lO 5.4 X 10-13 8.5 X 10-17 mol2 dm~ mol2 dm~ mol2 dm~ Solubility/ mol dm-3 1.34 X 10-5 7.35 X 10-7 9.22 X 10-9 • Note that: Solubility product Solubility Ksp (AgC/) > Ksp (AgBr) Solubility of AgC/ > Solubility of AgBr Ksp (AgBr) > Ksp (Agl) Solubility of AgBr > Solubility of Agl Ksp (AgC/) > Ksp (Ag2CrQ4) Solubility of AgC/ < Solubility of Ag2CrQ4 Units for Ks mol2 dm-6 Ag2CrQ4 1.1 X 10-12 mol3 dm-9 6.50 X 10-5 --------------------------------- - -------- - ------------- - ---------------------- - -----------ft-- The information above illustrate the following points: 0 Solubility products give a direct comparison of the solubility of two salts only if the total number of ions per unit formula of the compound produced in solution is the same in both cases, e.g. AgC/ and AgBr (or AgBr and Agl). In this case, the higher the Kso value, the higher the solubility of the ionic compound. 0 If the total number of ions produced is different, as in the case of AgC/ and Ag2CrQ4, then the solubility products may not give a direct comparison of the solubility of the two salts. For such cases, the solubility of the two salts should be used for comparison. -3-

2.3 Calculations of Solubility and Solubility Product • The solubility of a sparingly soluble salt at a given temperature can be determined experimentally and the data obtained can be used to calculate the solubility product of the salt. W:®111Exami>.lfil - Calculating Ksp from solubility The solubility of silver sulfide, Ag2S, is 2.48 x 10-15 mol dm-3 at 25 °C. Calculate its solubility product. eqm cone / mol dm-3 2(2.48 X 10-15) = 4.96 X 10-15 Ksp = [Ag•]2[S2-] = (4.96 x 10-15)2(2.48 x 10-15) = 6.10 x 1 Q-44 mol3 dm-9 - Calculating Ksp from solubility + 2.48 X 10-15 REMARKS • Write down the eqm equation. • Ag2S : Ag• : s2- = 1 : 2 : 1 • (Ag•] = 2[S2-J • ~P of a sparingly soluble compound is very small. The pH of a saturated solution of Fe(OH)2 is found to be 9.50 at 25 °C. Calculate the solubility product of Fe(OH)2 at 25 °C. pOH = 14- pH= 14-9.50 = 4.50 wM4Jtremii!il§ - Calculating solubility from Ksp REMARKS • Recall that pH+ pOH = 14 at 25 °C. • Write down the eqm reaction. • Fe(OH)2: Fe2• : OH- = 1 : 1 : 2 • [Fe2•] = ½ [OH-J Calculate the solubility of Mg(OH)2 in water if the value of its Ksp at 25 °C is 6.3 x 10-10. Let s mol dm-3 be the solubility of Mg(OH)2 in water. Mg(OH)2(s) .== Mg2•(aq) + 20H-(aq) eqm cone/ mol dm-J 5 ),; s (2/2),, Ur 1 ,,, -\) ..,,~ <; - \ ,40'f LJ / -t ~) Solubility of Mg(OH)2 in water at 25 °C = ( .l{Ji\ l v,A. l -4- REMARKS • When calculating solubility, "s" is usually used as the notation. • Write down the eqm reaction. • [Mg2•J = ½ [OH-J

3 I ionic Product and Precipitation L.,J 3.1 Ionic Product Consider a sparingly soluble ionic compound, MaXb(s): MaXb(S) aMb•(aq) + bXr(aq) The ionic product is the product of the concentration of the constituent ions in the solution at that instant raised to the powers of the stoichiometric coefficients: Ionic product= [Mb•(aq)]3[)(r(aq)]b :-Note=-----------------------------------------------------------------------------------------9. I :0 For Ksp, [Mb+(aq)] and [Xr(aq)] are equilibrium concentrations of Mb• and Xr in the