 # Solubility. Solubility “Insoluble” salts are governed by equilibrium reactions, and are really sparingly soluble. There is a dynamic equilibrium between.

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Solubility

Solubility “Insoluble” salts are governed by equilibrium reactions, and are really sparingly soluble. There is a dynamic equilibrium between the solvated ions and the crystalline solid.

Solubility

Solubility The extent to which a substance dissolves is its solubility. Solubility, as with all equilibria, is highly temperature dependent. Solubility may be expressed as grams of compound/100 mL of H 2 O, or as molarity, M.

Solubility Conventions The equilibrium constants for solubility are based on the insoluble solid appearing on the left side of the reaction, with the solvated ions on the right side. PbI 2 (s) ↔ Pb 2+ (aq) + 2I - (aq) K sp = [Pb 2+ ][ I - ] 2 = 1.4 x 10 -8 @25 o C sp stands for solubility product.

Solubility The molar solubility can be calculated from the value of K sp (and vice versa). Since salts have differing numbers of cations and anions, you cannot look at a table of K sp values and determine which compounds have the greater solubility.

K sp Values

Problem Calcium phosphate has a K sp of 1.3 x 10 -32. Calculate the molar solubility of calcium phosphate. Calcium phosphate has a K sp of 1.3 x 10 -32. Calculate the molar solubility of calcium phosphate.

Predicting Precipitation Reactions If the ions which may form a precipitate come from two different sources (the cation comes from one soluble salt, the anion from another), calculate the value of Q.

Predicting Precipitation Reactions Will a precipitate form if 10.0 mL of 0.010M AgNO 3 is added to 20.0 mL of 0.10M Na 2 SO 4 ? Will a precipitate form if 10.0 mL of 0.010M AgNO 3 is added to 20.0 mL of 0.10M Na 2 SO 4 ? 1. Determine the “insoluble” product. 2. Calculate the concentration of each ion of the “insoluble” product. 3. Calculate Q and compare the value to K sp.

pH and Solubility Since most anions are weak bases, solubility is highly pH dependent. The basic anion may become protonated in acidic solutions, and thus the “insoluble” salt will dissolve.

pH and Solubility For example, consider the solubility of CuS. K sp =8.5 x 10 -45. Sufide ion is the conjugate base of HS -, a weak acid with a K a of 1.3 x 10 -13. This means that sulfide ion is a relatively strong weak base, and will accept protons.

pH and Solubility Solubility Reaction: CuS(s) ↔ Cu 2+ (aq) + S 2- (aq) K sp = 8.5 x 10 -45 Reaction of Sulfide with Acid: S 2- (aq) + H 3 O + (aq) ↔ HS - (aq) + H 2 O(l) K b = K w /K a = (1.0 x 10 -14 )/1.3 x 10 -13 = 7.7 x 10 -2 Since sulfide is basic, CuS will dissolve in acidic solution.

pH and Solubility Since sulfide is basic, CuS will dissolve in acidic solution. As acid is added, this reaction is shifted towards the right. S 2- (aq) + H 3 O + (aq) ↔ HS - (aq) + H 2 O(l)

pH and Solubility Since sulfide is basic, CuS will dissolve in acidic solution. As acid is added, this reaction is shifted towards the right. S 2- (aq) + H 3 O + (aq) ↔ HS - (aq) + H 2 O(l)

pH and Solubility Since sulfide is basic, CuS will dissolve in acidic solution. As acid is added, this reaction is shifted towards the right. S 2- (aq) + H 3 O + (aq) ↔ HS - (aq) + H 2 O(l) The decrease in [S 2- ] shifts this reaction toward the right so as to make more sulfide. CuS(s) ↔ Cu 2+ (aq) + S 2- (aq)

pH and Solubility CuS(s) ↔ Cu 2+ (aq) + S 2- (aq) The result is a large increase in solubility of CuS in acidic solutions.

Problem: pH and Solubility Calculate the solubility of Fe(OH) 2 in water and in buffers of pH=3.00 and pH= 11.00. K sp for Fe(OH) 2 =1.8 x 10 -15. Calculate the solubility of Fe(OH) 2 in water and in buffers of pH=3.00 and pH= 11.00. K sp for Fe(OH) 2 =1.8 x 10 -15.

Selective Precipitation Mixtures of metal ions in aqueous solution are often separated by selective precipitation. A solution containing an anion that will only precipitate one of the metals while leaving the others in solution is used.

Selective Precipitation A solution contains 0.25M Ni(NO 3 ) 2 and 0.25M Cu(NO 3 ) 2. Can the metal ions be separated by slowly adding Na 2 CO 3 ? (Assume that for successful precipitation 99% of the metal ion must be precipitated before the other metal ion begins to precipitate, and assume no volume change on addition of sodium carbonate.) A solution contains 0.25M Ni(NO 3 ) 2 and 0.25M Cu(NO 3 ) 2. Can the metal ions be separated by slowly adding Na 2 CO 3 ? (Assume that for successful precipitation 99% of the metal ion must be precipitated before the other metal ion begins to precipitate, and assume no volume change on addition of sodium carbonate.)

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