Analytical Chem istry - DePauw University

236 Analytical Chemistry 2.0When we first add solid Pb(IO 3 ) 2 to water,the concentrations of Pb 2+ and IO 3–are zero and the reaction quotient, Q, isQ = [Pb 2+ ][IO 3 – ] 2 = 0As the solid dissolves, the concentrationsof these ions increase, but Q remainssmaller than K sp . We reach equilibriumand “satisfy the solubility product” whenQ = K spBecause a solid, such as Pb(IO 3 ) 2 , doesnot appear in the solubility product expression,we do not need to keep track ofits concentration. Remember, however,that the K sp value applies only if there issome Pb(IO 3 ) 2 present at equilibrium.6G.1 A Simple Problem—Solubility of Pb(IO 3 ) 2If we place an insoluble compound such as Pb(IO 3 ) 2 in deionized water, thesolid dissolves until the concentrations of Pb 2+ and IO 3 – satisfy the solubilityproduct for Pb(IO 3 ) 2 . At equilibrium the solution is saturated withPb(IO 3 ) 2 , which simply means that no more solid can dissolve. How dowe determine the equilibrium concentrations of Pb 2+ and IO 3 – , and whatis the molar solubility of Pb(IO 3 ) 2 in this saturated solution?We begin by writing the equilibrium reaction and the solubility productexpression for Pb(IO 3 ) 2 .Pb(IO ) () s Pb 2 + ( ) −aq + 2IO( aq )3 2 32+ − 2 −13K sp= [ Pb ][ IO ] = 25 . × 10 6.333As Pb(IO 3 ) 2 dissolves, two IO 3 – ions are produced for each ion of Pb 2+ . Ifwe assume that the change in the molar concentration of Pb 2+ at equilibriumis x, then the change in the molar concentration of IO 3 – is 2x. Thefollowing table helps us keep track of the initial concentrations, the changein concentrations, and the equilibrium concentrations of Pb 2+ and IO 3 – .Concentrations Pb(IO 3 ) 2 (s) Pb 2+ (aq) + 2IO 3– (aq)Initial solid 0 0Change solid +x +2xEquilibrium solid x 2xSubstituting the equilibrium concentrations into equation 6.33 and solvinggives( x)( 2x) = 4x= 25 . × 102 3 −13x = 397 . × 10Substituting this value of x back into the equilibrium concentration expressionsfor Pb 2+ and IO 3 – gives their concentrations as−5We can express a compound’s solubilityin two ways: molar solubility (mol/L) ormass solubility (g/L). Be sure to expressyour answer clearly.[ Pb ] = x = 40 . × 102+ −5−[IO ] = 2x= 7.9×103Because one mole of Pb(IO 3 ) 2 contains one mole of Pb 2+ , the molar solubilityof Pb(IO 3 ) 2 is equal to the concentration of Pb 2+ , or 4.0 × 10 –5 M.Practice Exercise 6.7Calculate the molar solubility and the mass solubility for Hg 2 Cl 2 , giventhe following solubility reaction and K sp value.−5Hg Cl () s Hg ( aq) + 2Cl( aq ) K = 1.2×102 18 + − −2 2 2spClick here to review your answer to this exercise.MM