Section 4: Solubility Equilibrium

Objectives Explain what is meant by solubility product constants, and calculate their values. Calculate solubilities using solubility product constants. Carry out calculations to predict whether precipitates will form when solutions are combined.

Solubility Product A saturated solution contains the maximum amount of solute possible at a given temperature in equilibrium with an undissolved excess of the substance. A saturated solution is not necessarily a concentrated solution. The equilibrium principles developed in this chapter apply to all saturated solutions of sparingly soluble salts.

Solubility Product, continued The heterogeneous equilibrium system in a saturated solution of silver chloride containing an excess of the solid salt is represented by The solubility product constant, K sp, of a substance is the product of the molar concentrations of its ions in a saturated solution, each raised to the power that is the coefficient of that ion in the balanced chemical equation.

Solubility Product, continued The equation for the solubility equilibrium expression for the dissolution reaction of AgCl is The equilibrium expression is written without including the solid species. The numerical value of K sp can be determined from solubility data.

Solubility Product, continued For a saturated solution of CaF 2, the equilibrium equation is The expression for the solubility product constant is The solubility of CaF 2 is is 8.6  10 −3 /100 g of water at 25°C. Expressed in moles per liter this concentration becomes 1.1  10 −3 mol/L.

Determining Ksp for Reactions at Chemical Equilibrium

Solubility Product, continued CaF 2 dissociates to yield twice as many F − ions as Ca 2+ ions. [Ca 2+ ] = 1.1  10 −3 mol/L [F − ] = 2.2  10−3 mol/L K sp = 5.3  Calculations of K sp ordinarily should be limited to two significant figures.

Solubility Product Constants at 25°C

Solubility Product, continued The solubility product constant is an equilibrium constant representing the product of the molar concentrations of its ions in a saturated solution. The solubility of a solid is an equilibrium position that represents the amount of the solid required to form a saturated solution with a specific amount of solvent. It has only one value for a given solid at a given temperature. It has an infinite number of possible values at a given temperature and is dependent on other conditions, such as the presence of a common ion.

Solubility Product, continued Sample Problem B Calculate the solubility product constant, K sp,for copper(I) chloride, CuCl, given that the solubility of this compound at 25°C is 1.08  10 –2 g/100. g H 2 O.

Solubility Product, continued Sample Problem B Solution K sp =[Cu + ][Cl – ] Unknown: K sp Given: solubility of CuCl = 1.08  10 −2 g CuCl/100. g H 2 O Solution: [Cu + ] = [Cl – ] = solubility in mol/L

Solubility Product, continued Sample Problem B Solution, continued 1.09  mol/L CuCl K sp = (1.09  )(1.09  ) = 1.19  [Cu + ] = [Cl – ]=1.09  mol/L

Calculating Solubilities The solubility product constant can be used to determine the solubility of a sparingly soluble salt. The molar solubility of BaCO 3 is 7.1  10 −5 mol/L. How many moles of barium carbonate, BaCO 3, can be dissolved in 1 L of water at 25°C?

Calculating Solubilities, continued Sample Problem C Calculate the solubility of silver bromide, AgBr, in mol/L, using the K sp value for this compound.

Calculating Solubilities, continued Sample Problem C Solution Given: K sp = 5.0  10 −13 Unknown: solubility of AgBr Solution: [Ag + ] = [Br − ], so let [Ag + ] = x and [Br − ] = x

Calculating Solubilities, continued Sample Problem C Solution, continued

Precipitation Calculations The equilibrium condition does not require that the two ion concentrations be equal. Equilibrium will still be established so that the ion product does not exceed the value of K sp for the system. If the ion product is less than the value of K sp at a particular temperature, the solution is unsaturated. If the ion product is greater than the value for K sp, solid precipitates.

Precipitation Calculations, continued Unequal quantities of BaCl 2 and Na 2 CO 3 are dissolved in water and the solutions are mixed. If the ion product exceeds the K sp of BaCO 3, a precipitate of BaCO 3 forms. Precipitation continues until the ion concentrations decrease to the point at which equals the K sp. The solubility product can be used to predict whether a precipitate forms when two solutions are mixed.

Precipitation Calculations, continued Sample Problem D Will a precipitate form if 20.0 mL of M BaCl 2 is mixed with 20.0 mL of M Na 2 SO 4 ?

Precipitation Calculations, continued Sample Problem D Solution Given: concentration of BaCl 2 = M volume of BaCl 2 = 20.0 mL concentration of Na 2 SO 4 = M volume of Na 2 SO 4 = 20.0 mL Unknown: whether a precipitate forms Solution: The two possible new pairings of ions are NaCl and BaSO 4. BaSO 4 is a sparingly soluble salt.

Precipitation Calculations, continued Sample Problem D Solution, continued mol Ba 2+ ion:

Precipitation Calculations, continued Sample Problem D Solution, continued total volume of solution: L L = L concentration Ba 2+ ion in combined solution:

Precipitation Calculations, continued Sample Problem D Solution, continued The ion product: Precipitation occurs.

Limitations on the Use of K sp The solubility product principle can be very useful when applied to solutions of sparingly soluble substances. It cannot be applied very successfully to solutions of moderately soluble or very soluble substances. The positive and negative ions attract each other, and this attraction becomes appreciable when the ions are close together. Sometimes it is necessary to consider two equilibria simultaneously.

Equilibrium Calculations