Presentation on theme: "Chapter 15 Solubility Equilibria AP Chemistry. The Solubility Product Constant, K sp Many important ionic compounds are only slightly soluble in water."— Presentation transcript:
Chapter 15 Solubility Equilibria AP Chemistry
The Solubility Product Constant, K sp Many important ionic compounds are only slightly soluble in water and equations are written to represent the equilibrium between the compound and the ions present in a saturated aqueous solution. The solubility product constant, K sp, is the product of the concentrations of the ions involved in a solubility equilibrium, each raised to a power equal to the stoichiometric coefficient of that ion in the chemical equation for the equilibrium.
The Solubility Equilibrium Equation And K sp CaF 2 (s) Ca 2+ (aq) + 2 F - (aq) K sp = [Ca 2+ ][F - ] 2 K sp = 5.3x10 -9 As 2 S 3 (s) 2 As 3+ (aq) + 3 S 2- (aq) K sp = [As 3+ ] 2 [S 2- ] 3 * Remember, solids are not in equilibrium expressions!
Some Values For Solubility Product Constants (K sp ) At 25 o C
K sp And Molar Solubility The solubility product constant (K sp ) is related to the solubility of an ionic solute, but K sp and Molar Solubility - the molarity of a solute in a saturated aqueous solution Ksp and Molar solubility are not the same thing. Calculating solubility equilibria fall into two categories: –determining a value of K sp from experimental data –calculating equilibrium concentrations when K sp is known.
Calculating K sp From Molar Solubility It is found that 1.2x10 -3 mol of lead (II) iodide, PbI 2, dissolves in 1.0 L of aqueous solution at 25 o C. What is the K sp at this temperature? Ksp = [Pb 2+ ][I - ] 2 PbI 2 Pb I - Ksp = [1.2 x ] [2(1.2 x )] 2 2 iodide ions form, so you must multiply molarity by 2! Ksp = 6.9 x 10 -9
Calculating Molar Solubility From K sp Calculate the molar solubility of silver chromate, Ag 2 CrO 4, in water from K sp = 1.1x for Ag 2 CrO 4. Ag 2 CrO 4 (s) 2Ag + + CrO 4 2- Ksp = [Ag + ] 2 [CrO 4 -2 ] At equilibrium 2x x 1.1 x = (x)(2x) x = 4x 3 x = 6.5 x M = [Ag 2 CrO 4 ] = [CrO 4 2- ] [Ag + ] = 2(6.5 x )
The Common Ion Effect In Solubility Equilibria The common ion effect also affects solubility equilibria. Le Châtelier’s principle is followed for the shift in concentration of products and reactants upon addition of either more products or more reactants to a solution. The solubility of a slightly soluble ionic compound is lowered when a second solute that furnishes a common ion is added to the solution.
Solubility Equilibrium Calculation -The Common Ion Effect What is the solubility of Ag 2 CrO 4 in 0.10 M K 2 CrO 4 ? K sp = 1.1x for Ag 2 CrO 4. Comparison of solubility of Ag 2 CrO 4 In pure water:6.5 x M In 0.10 M K 2 CrO 4 :1.7 x M The common ion effect!! Adding more CrO 4 2- ions shifts the equilibrium back to the reactants, which is solid Ag 2 CrO 4 Ag 2 CrO 4 2Ag + + CrO 4 2-
Determining Whether Precipitation Occurs Q sp is the ion product reaction quotient and is based on initial conditions of the reaction. Q sp can then be compared to K sp. To predict if a precipitation occurs: - Precipitation should occur if Q sp > K sp. - Precipitation cannot occur if Q sp < K sp. - A solution is just saturated if Q sp = K sp. Sometimes the concentrations of the ions are not high enough to produce a precipitate!
Determining Whether Precipitation Occurs – An Example The concentration of calcium ion in blood plasma is M. If the concentration of oxalate ion is 1.0x10 -7 M, do you expect calcium oxalate to precipitate? K sp = 2.3x Three steps: (1)Determine the initial concentrations of ions. (2)Evaluate the reaction quotient Q ip. (3)Compare Q sp with K sp.
Determining Whether Precipitation Occurs In applying the precipitation criteria, the effect of dilution when solutions are mixed must be considered. Example: A mL sample of M Pb(NO 3 ) 2 (aq) is mixed with mL of M NaI (aq). Should precipitation of PbI 2 (s), K sp = 7.1x10 -9, occur? Calculate new concentrations in total volume of 400mls = 0.4L [Pb 2+ ] = (0.250L)(0.0012M)/(0.400L) = 7.5 x M [I-] = (0.150L)(0.0640M)/(0.400L) = M Qsp = [Pb 2+ ][I - ] 2 = (7.5 x )(0.024) 2 = 4.32 x Qsp > Ksp therefore a precipitate will form!
Selective Precipitation a)The first precipitate to form when AgNO 3 (aq) is added to an aqueous solution containing Cl - and I - is yellow AgI(s). b)Essentially all the I - has precipitated before the precipitation of white AgCl(s) begins. AgCl (s) Ag+ (aq) + Cl- (aq) Ksp = 1.8x AgI (s) Ag+ (aq) + I- (aq) Ksp = 8.5x10 -17
Selective Precipitation An Example Example: An aqueous solution that is 2.00 M in AgNO 3 is slowly added from a buret to an aqueous solution that is M in Cl - and also M in I - AgCl (s) Ag + (aq) + Cl - (aq)K sp = 1.8x AgI (s) Ag + (aq) + I - (aq)K sp = 8.5x a.Which ion. Cl - or I -, is the first to precipitate from solution? (this depends on the Molar solubility not Ksp, but since both compounds give equal ions, then you can use Ksp) a.Is separation of the two ions by selective precipitation feasible? AgI will precipitate first, lower Ksp
Effect of pH on Solubility The solubility of an ionic solute may be greatly affected by pH if an acid-base reaction also occurs as the solute dissolves. In other words, some salts will not dissolve well in pure water, but will dissolve in an acid or a base. If the anion (A-) of the salt/precipitate is that of a weak acid, the salt/precipitate will dissolve more when in a strong acid (H+ ions will form HA with A-) However, if the anion of the precipitate is that of a strong acid, adding a strong acid will have no effect on the precipitate dissolving more.
Qualitative Inorganic Analysis Acid-base chemistry, precipitation reactions, oxidation-reduction, and complex-ion formation all come into sharp focus in an area of analytical chemistry called classical qualitative inorganic analysis. “Qualitative” signifies that the interest is in determining what is present, not how much is present. Although classical qualitative analysis is not as widely used today as instrumental methods, it is still a good vehicle for applying all the basic concepts of equilibria in aqueous solutions.
Cations of Group 1 If aqueous HCl is added to an unknown solution of cations, and a precipitate forms, then the unknown contains one or more of these cations: Pb 2+, Hg 2 2+, or Ag +. These are the only ions to form insoluble chlorides. If there is no precipitate, then these ions must be absent from the mixture. If there is a precipitate, it is filtered off and saved for further analysis. The supernatant liquid is also saved for further analysis.
Cation Group 1 (continued) Analyzing For Pb 2+ Of the three possible ions in solution, PbCl 2 is the most soluble in water. The precipitate is washed with hot water and the washings then treated with aqueous K 2 CrO 4. If Pb 2+ is present, chromate ion combines with lead ion to form a precipitate of yellow lead chromate, which is less soluble than PbCl 2. If Pb 2+ is absent, then the washings just become tinged yellow but no precipitate is in evidence.
Cation Group 1 (continued) Analyzing For Ag + Next, the undissolved precipitate is treated with aqueous ammonia. If AgCl is present, it will dissolve in this solution. If there is any remaining precipitate, it is separated from the supernatant liquid and saved for further analysis. The supernatant liquid (which contains the Ag +, if present) is then treated with aqueous nitric acid. If a precipitate reforms, then Ag + was present in the solution, if no precipitate forms, then Ag + was not present in the solution.
Cation Group 1 (continued) Analyzing For Hg 2 2+ When precipitate was treated with aqueous ammonia in the previous step, any Hg 2 2+ underwent an oxidation-reduction reaction to form a dark gray mixture of elemental mercury and HgNH 2 Cl that precipitates from the solution. If this dark gray precipitate was observed, then mercury was present in the original unknown sample. If this dark gray precipitate was not observed, then mercury must have been absent from the original unknown sample.
Group 1 Cation Precipitates left: cation goup 1 ppt: PbCl 2, PbCl 2, AgCl (all white) middle: product from test for Hg 2 2+ : mix of Hg (black) and HgNH 2 Cl (white) right: product from test for Pb 2+ : PbCrO 4 (yellow) when K 2 CrO 4(aq) is reacted with saturated PbCl 2