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

SolubilitySolubility


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


The extent to which a substance The extent to which a substance dissolves is its solubility. Solubility, dissolves is its solubility. Solubility, as with all equilibria, is highly as with all equilibria, is highly temperature dependent. temperature dependent.

Solubility may be expressed as Solubility may be expressed as grams of compound/100 mL of Hgrams of compound/100 mL of H22O, O, or as molarity, M.or as molarity, M.

Solubility ConventionsSolubility Conventions

The equilibrium constants for The equilibrium constants for solubility are based on the insoluble solubility are based on the insoluble solid appearing on the left side of the solid appearing on the left side of the reaction, with the solvated ions on the reaction, with the solvated ions on the right side.right side.

PbIPbI22((ss) ↔ Pb) ↔ Pb2+2+((aqaq) + 2I) + 2I--((aqaq))

KKspsp = [Pb = [Pb2+2+][ I][ I- - ]]22 = 1.4 x 10 = 1.4 x 10-8-8 @25 @25ooCC

sp stands for solubility product.sp stands for solubility product.


The molar solubility can be The molar solubility can be calculated from the value of Kcalculated from the value of Kspsp (and (and vice versa). vice versa).

Since salts have differing Since salts have differing numbers of cations and anions, you numbers of cations and anions, you cannot look at a table of Kcannot look at a table of Kspsp values values and determine which compounds and determine which compounds have the greater solubility.have the greater solubility.

KKspsp Values Values

ProblemProblem

Calcium phosphate has a KCalcium phosphate has a Kspsp of 1.3 x of 1.3 x 1010-32-32. Calculate the molar solubility . Calculate the molar solubility of calcium phosphate.of calcium phosphate.

Predicting Precipitation Predicting Precipitation ReactionsReactions

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

Predicting Precipitation Predicting Precipitation ReactionsReactions

Will a precipitate form if 10.0 mL of Will a precipitate form if 10.0 mL of 0.010M AgNO0.010M AgNO33 is added to 20.0 mL of is added to 20.0 mL of 0.10M Na0.10M Na22SOSO44??

1.1. Determine the “insoluble” product.Determine the “insoluble” product.

2.2. Calculate the concentration of each ion Calculate the concentration of each ion of the “insoluble” product.of the “insoluble” product.

3.3. Calculate Q and compare the value to Calculate Q and compare the value to KKspsp..

pH and SolubilitypH and Solubility

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


For example, consider the For example, consider the solubility of CuS. Ksolubility of CuS. Kspsp=8.5 x 10=8.5 x 10-45-45. . Sufide ion is the conjugate base of Sufide ion is the conjugate base of HSHS--, a weak acid with a K, a weak acid with a Kaa of 1.3 x of 1.3 x 1010-13-13. This means that sulfide ion is a . This means that sulfide ion is a relatively strong weak base, and will relatively strong weak base, and will accept protons.accept protons.


Solubility Reaction:Solubility Reaction:

CuS(CuS(ss) ↔ Cu) ↔ Cu2+2+((aqaq) + S) + S2-2-((aqaq) K) Kspsp= 8.5 x 10= 8.5 x 10-45-45

Reaction of Sulfide with Acid:Reaction of Sulfide with Acid:

SS2-2-((aqaq) + H) + H33OO++((aqaq) ↔ HS) ↔ HS--((aqaq) + H) + H22O(O(ll))

KKbb = K = Kww/K/Kaa = (1.0 x 10 = (1.0 x 10-14-14)/1.3 x 10)/1.3 x 10-13-13 = 7.7 x 10 = 7.7 x 10--

22

Since sulfide is basic, CuS will dissolve in Since sulfide is basic, CuS will dissolve in acidic solution.acidic solution.


Since sulfide is basic, CuS will Since sulfide is basic, CuS will dissolve in acidic solution. As acid is dissolve in acidic solution. As acid is added, this reaction is shifted added, this reaction is shifted towards the right.towards the right.



Since sulfide is basic, CuS will Since sulfide is basic, CuS will dissolve in acidic solution. As acid is dissolve in acidic solution. As acid is added, this reaction is shifted towards the added, this reaction is shifted towards the right.right.


The decrease in [SThe decrease in [S2-2-] shifts this ] shifts this reaction toward the right so as to make reaction toward the right so as to make more sulfide.more sulfide.

CuS(CuS(ss) ↔ Cu) ↔ Cu2+2+((aqaq) + S) + S2-2-((aqaq))


CuS(CuS(ss) ↔ Cu) ↔ Cu2+2+((aqaq) + S) + S2-2-((aqaq))

The result is a large increase in The result is a large increase in solubility of CuS in acidic solutions. solubility of CuS in acidic solutions.

Problem: pH and Problem: pH and SolubilitySolubility

Calculate the solubility of Fe(OH)Calculate the solubility of Fe(OH)22 in in water and in buffers of pH=3.00 and water and in buffers of pH=3.00 and pH= 11.00. KpH= 11.00. Kspsp for Fe(OH) for Fe(OH)2 2 =1.8 x =1.8 x 1010-15-15..

Selective PrecipitationSelective Precipitation

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

Selective PrecipitationSelective Precipitation

A solution contains 0.25M Ni(NOA solution contains 0.25M Ni(NO33))22 and 0.25M Cu(NOand 0.25M Cu(NO33))22. Can the metal . Can the metal ions be separated by slowly adding ions be separated by slowly adding NaNa22COCO33? (Assume that for successful ? (Assume that for successful precipitation 99% of the metal ion precipitation 99% of the metal ion must be precipitated before the other must be precipitated before the other metal ion begins to precipitate, and metal ion begins to precipitate, and assume no volume change on addition assume no volume change on addition of sodium carbonate.)of sodium carbonate.)