Solubility. Solubility. 13-Jun-11. Chem 1011 Intersession 2011 Class #32. Solubility Equilibria / Solubility Product Constant (K sp ) 1

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1 1 2 Class 32: Solubility Equilibria / Solubility Product (K sp ) Sec 16.5 Solubility Equilibria and Solubility Product Constant K sp and Molar Solubility K sp and Relative Solubility The Effect of a Common Ion on Solubility The Effect of ph on Solubility Sec 16.5 Solubility Soluble a compound dissolves in water Insoluble a compound does not dissolve in water. Note: All ionic compounds dissolve in water to some degree, however, many compounds have such low solubility in water that we classify them as insoluble There is a continuum of solubility, and we are now going to examine degrees of solubility 3 4 Solubility In Chem 1010 you learned solubility rules. Please review these; it is important to know what compounds are always soluble. (See Tro, Table 14.1) Solubility Equilibria and Solubility Product Constant (K sp ) Solubility Product constant (K sp ) equilibrium expression for the dissolution of an ionic compound. Example CaF 2(s) Ca 2+ (aq) + 2F (aq) K sp = 1.46 x K sp = [Ca 2+ ][F ] 2 Just like any other equilibrium, solids do not appear in K expressions just the aqueous ions! The value of K sp is a measure of the solubility of a compound 5 6 Solubility Product Constant (K sp ) K sp and Molar Solubility Molar solubility the solubility in units of moles per litre (mol/l) Note K sp is NOT molar solubility; K sp can be used to determine molar solubility. K sp has only one value for every temperature. Molar solubility can have different values, depending on the solution conditions (common ion effect, etc ). Molar solubility is normally given the symbol s (to represent solubility) when working with K sp ICE tables. Constant (K sp ) 1

2 7 8 K sp and Molar Solubility Example: Calculate the molar solubility of PbCl 2 in pure water at 25 C. (K sp = 1.17 x 10-5 ) Write the dissociation reaction and K sp expression. Create an ICE table defining the change in terms of the solubility of the solid. K sp = [Pb 2+ ][Cl ] 2 [I] 0 0 [C] +s +2s [E] s 2s Ksp and Molar Solubility Substitute into the K sp expression. Substitute the value of K sp into the equation, and solve for s. K sp = [Pb 2+ ][Cl ] 2 K sp = (s)(2s) s Calculate the molar solubility of Fe(OH) 2 in pure water. K sp (Fe(OH) 2 ) = 4.87 x K sp and Relative Solubility You cannot always use K sp values to directly compare the molar solubility of two compounds. The molar solubility of AgBr in pure water is 7.3 x 10-7 M. Calculate K sp. This is because the relationship between K sp and molar solubility also depends on the stoichiometry of the dissociation process. If two reactions have the same dissociation stoichiometry, their molar solubilities can be directly compared The solubility of an ionic compound is lower in a solution containing a common ion than in pure water. This makes sense, based upon Le Chatelier s principle. Example: Consider a solution of PbCl 2, to which mol of NaCl is added. left. Increase the concentration of Cl Example: Calculate the molar solubility of CaF 2 in M NaF at 25 C. (Ksp = 1.46 x ) Write the dissociation reaction and K sp expression. Create an ICE table defining the change in terms of the solubility of the solid. CaF 2(s) Ca 2+ (aq) + 2 F (aq) CaF 2(s) K sp = [Ca 2+ ][F ] 2 Ca 2+ (aq) + 2 F (aq) [I] M [C] +s +2s [E] s s Constant (K sp ) 2

3 13 14 Substitute into the K sp expression. Substitute the value of K sp into the equation, and solve for s. K sp = [Ca 2+ ][F ] 2 K sp = (S)( S) 2 Since K sp is small, assume s is small So, s K sp = (S)(0.100) 2 Calculate the molar solubility of CaF 2 in a solution containing M Ca(NO 3 ) 2 (K sp (CaF 2 ) = 1.46 x ) The Effect of ph on Solubility The solubility of an ionic compound with a strongly or weakly basic anion increases with increasing acidity (decreasing ph). This makes sense, based upon Le Chatelier s principle. Example: Consider a solution of Mg(OH) 2, in basic media (high ph; high [OH ].) Mg(OH) 2(s) Mg 2+ (aq) + 2 OH (aq) left. High [OH ] The Effect of ph on Solubility The solubility of an ionic compound with a strongly or weakly basic anion increases with increasing acidity (decreasing ph). Again, this makes sense based on LeChatelier s principle. Example: Consider a solution of Mg(OH) 2, in acidic media (low ph; OH being neutralized by H 3 O +.) Mg(OH) 2(s) Mg 2+ (aq) + 2 OH (aq) right. H 3 O + reacts with OH ; effectively, OH is being removed from the system Which compound, FeCO 3 or PbBr 2, is more soluble in acid than in base? Why? will occur upon the mixing of two solutions containing ionic compounds if the cross product (combination of the cation from one solution and anion from the other) is insoluble. We need to quantify insoluble in order to determine if precipitation will take place. Constant (K sp ) 3

4 Chem 1011 Dr. L. Dawe Winter (Soluble) ( Insoluble ) If we compare the reaction quotient, Q, for the current solution concentrations to the value of K sp, we can determine whether precipitation will occur. Q = K sp : the solution is saturated; no precipitation Q < K sp : the solution is unsaturated; no precipitation Q > K sp : the solution is above saturation; the salt above saturation will precipitate Will a precipitate form when we mix Pb(NO 3 ) 2(aq) with NaBr (aq) if the concentrations after mixing are M and M, respectively? 1. Write an equation for the reaction Pb(NO 3 ) 2(aq) + 2 NaBr (aq) PbBr 2(s) + 2 NaNO 3(aq) 2. Determine the ion concentrations of the original salts Pb(NO 3 ) 2 = M NaBr = M Pb 2+ = M Na + = M NO 3 = 2( M) Br = M Precepitation 3. Determine the K sp for any insoluble products In the equation, PbBr 2 is a solid product In a test or exam, K sp values are provided when practicing, there is a list in your textbook K sp (PbBr 2 ) = 4.67 x Write the dissociation equation for the solid: PbBr 2(s) Pb 2+ (aq) + 2 Br (aq) Write the expression for Q sp, and substitute the ion concentrations to solve A solution containing M Pb(NO 3 ) 2 and a solution containing M NaBr are mixed. Will a precipitate form? (Ksp (PbBr 2 ) = 4.67 x 10-6 ) 6. Compare Q sp to K sp and determine the direction of the shift Q sp = 1.84 x 10 7 < K sp = 4.67 x 10 6 Since Q sp < K sp, the equilibrium shifts right (towards products; away from the solid) and so no precipitate forms Constant (K sp ) 4

5 25 Last Chapter! Sec 18.3 Voltaic (or Galvanic) Cells: Generating Electricity from Spontaneous Chemical Reactions Electrochemical Cell Notation Sec 18.4 Standard Electrode Potentials Predicting the Spontaneous Direction of an Oxidation-Reduction Reaction Predicting Whether a Metal will Dissolve in Acid Constant (K sp ) 5