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.
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.
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, Ksp, 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.
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 Kspcan be determined from solubility data.
For a saturated solution of CaF2, the equilibrium equation is
The expression for the solubility product constant is
The solubility of CaF2 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
The solubility product constant can be used to determine the solubility of a sparingly soluble salt.
How many moles of barium carbonate, BaCO3, can be dissolved in 1 L of water at 25°C?
The molar solubility of BaCO3 is 7.1 10−5 mol/L.
Sample Problem C Calculate the solubility of silver bromide, AgBr, in mol/L, using the Kspvalue for this compound. Given:Ksp= 5.0 10−13 Unknown:solubility of AgBr Solution: [Ag+] = [Br−], so let [Ag+] = x and [Br−] = x
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 Kspfor the system.
If the ion product is less than the value of Kspat a particular temperature, the solution is unsaturated.
If the ion product is greater than the value for Ksp, solid precipitates.
Unequal quantities of BaCl2 and Na2CO3 are dissolved in water and the solutions are mixed.
If the ion product exceeds the Kspof BaCO3, a precipitate of BaCO3 forms.
Precipitation continues until the ion concentrations decrease to the point at which equals the Ksp.
The solubility product can be used to predict whether a precipitate forms when two solutions are mixed.
Sample Problem D Will a precipitate form if 20.0 mL of 0.010 M BaCl2is mixed with 20.0 mL of 0.0050 M Na2SO4? Given:concentration of BaCl2 = 0.010 M volume of BaCl2 = 20.0 mL concentration of Na2SO4 = 0.0050 M volume of Na2SO4 = 20.0 mL Unknown:whether a precipitate forms Solution: The two possible new pairings of ions are NaCl and BaSO4. BaSO4 is a sparingly soluble salt.