Analytical Chem istry - DePauw University

238 Analytical Chemistry 2.0the assumed concentration. This is a negligible error. Accepting the resultof our calculation, we find that the equilibrium concentrations of Pb 2+ andIO 3 – are2+[ Pb ] = 010 . + x ≈ 0.10 M−[IO ] = 2x= 16 . × 103The molar solubility of Pb(IO 3 ) 2 is equal to the additional concentration ofPb 2+ in solution, or 7.9 × 10 –4 mol/L. As expected, Pb(IO 3 ) 2 is less solublein the presence of a solution that already contains one of its ions. This isknown as the common ion effect.As outlined in the following example, if an approximation leads to anunacceptably large error we can extend the process of making and evaluatingapproximations.Example 6.10Calculate the solubility of Pb(IO 3 ) 2 in 1.0 × 10 –4 M Pb(NO 3 ) 2 .So l u t i o nLetting x equal the change in the concentration of Pb 2+ , the equilibriumconcentrations of Pb 2+ and IO 3 – are−6M2[ Pb+ 4] = 10 . × 10 − + x [ IO− ] = 2xSubstituting these concentrations into equation 6.33 leaves us with−(. 10× 10 + x)( 2x) = 2.5×104 2 −13To solve this equation for x, we make the following assumption3[ Pb ] = 10 . × 10 + x ≈ 10 . × 102 + − 4 − 4Mobtaining a value for x of 2.50× 10 –4 . Substituting back, gives the calculatedconcentration of Pb 2+ at equilibrium as[ Pb ] = 10 . × 10 + 250 . × 10 = 1.25×102+ −4 −5 −4Ma value that differs by 25% from our assumption that the equilibriumconcentration is 1.0× 10 –4 M. This error seems unreasonably large. Ratherthan shouting in frustration, we make a new assumption. Our first assumption—thatthe concentration of Pb 2+ is 1.0× 10 –4 M—was too small.The calculated concentration of 1.25× 10 –4 M, therefore, is probably a bittoo large. For our second approximation, let’s assume that[ ] = 10 . × 10 + x ≈ 125 . × 10Pb 2+ −4 −4Substituting into equation 6.33 and solving for x gives its value as2.24× 10 –5 . The resulting concentration of Pb 2+ is