chapter sixteenprentice-hall ©2002slide 1 of 32 solubility products heterogeneous equilibria...

Chapter SixteenPrentice-Hall ©2002 Slide 1 of 32

Solubility Products

Heterogeneous Equilibria Slightly Soluble Salts

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Solubility of Ionic Salts

The solubility of ionic salts varies considerably. Most nitrates are very soluble, while phosphates and may other salts of transition metal ions are often insoluble.

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Solubility and Temperature

The solubility of ionic salts varies considerably with the temperature. Some salts such as KNO3

have a wide solubility range. On the other hand, the solubility of NaCl in water varies only slightly.

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The Solubility Product Constant, Ksp

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, Ksp, 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.

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The Solubility Equilibrium Equation And Ksp

Example 1

CaF2 (s) Ca2+ (aq) + 2 F- (aq)

Ksp = [Ca2+][F-]2 Ksp = 5.3x10-9

Example 2As2S3 (s) 2 As3+ (aq) + 3 S2- (aq)

Ksp = [As3+]2[S2-]3 Ksp = 4.4 x10-27

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Some Values For Solubility Product Constants (Ksp) At 25 oC

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Ksp And Molar Solubility

The solubility product constant is related to the solubility of an ionic solute, but Ksp and molar solubility - the molarity of a solute in a saturated aqueous solution - are not the same thing.

Calculating solubility equilibria fall into two categories: determining a value of Ksp from experimental data calculating equilibrium concentrations when Ksp is

known.

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Calculating Ksp From Molar Solubility

It is found that 1.2x10-3 mol of lead (II) iodide, PbI2, dissolves in 1.0 L of aqueous solution at 25 oC. What is the Ksp at this temperature?

PbI2 (s) Pb2+ (aq) + 2 I- (aq)

Ksp = [Pb2+] [I-]2

For every lead iodide that dissolves there is one lead ion and 2 iodide ions. Therefore

[Pb2+] = 1.2x10-3

[I-] = 2.4x10-3

Ksp =(1.2x10-3)(2.4x10-3)2 = 6.9 x 10-9

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Calculating Molar Solubility From Ksp

Calculate the concentration of [Ag+] and [Cl-] in a saturated solution of silver chloride. The ksp for silver chloride : Ksp = 1.8x10-10

Ksp = [Ag+] [Cl-] = 1.8x10-10

Let x = [Ag+] = [Cl-] .

Substitute into the Ksp expression

x2 = 1.8x10-10 x = 1.8x10-10 =1.34 x 10-5

[Ag+] = [Cl-] =1.34 x 10-5

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Calculating Molar Solubility From Ksp

Calculate the molar solubility of silver chromate, Ag2CrO4, in water from Ksp = 1.1x10-12 for Ag2CrO4.

Ag2CrO4 (s) 2 Ag+ + CrO42-

Ksp = [Ag+]2 [CrO42-] = 1.1x10-12

Let x = [CrO42-] 2x = [Ag+]

Substitute into the Ksp expression

4x3 = 1.1x10-12 x = 3 1.1x10-12 /4

x = [CrO42-] = 6.5 x10-5

2x = [Ag+] = 1.30 x 10-4

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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.

The common ion effect is frequently used in analytical chemistry to determine such things as the percent of an element present in an unknown sample.

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Solubility Equilibrium Calculation-The Common Ion Effect

What is the solubility of Ag2CrO4 in 0.10 M K2CrO4? Ksp = 1.1x10-12 for Ag2CrO4.

Let 2x = Ag+ and CrO42- = 10-1

(4x2)(10-1) = 1.1 x 10-12

4x2 = 1.1 x 10-12/ 10-1 = 1.1 x 10-11 x = 1.66 x 10-6

[Ag+] = 2x = 3.32 x 10-6

Comparison of solubility of Ag2CrO4

In pure water: 6.5 x 10-5 MIn 0.10 M K2CrO4: 1.7 x 10-6 MThe common ion effect!!

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Determining Whether Precipitation Occurs

Qip is the ion product reaction quotient and is based on initial conditions of the reaction.

Qip can then be compared to Ksp. To predict if a precipitation occurs:

- Precipitation should occur if Qip > Ksp.

- Precipitation cannot occur if Qip < Ksp.

- A solution is just saturated if Qip = Ksp.

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Determining Whether Precipitation Occurs – Example 1

[Pb2+] = 2.00 x 10-3 so [I-]2 = (7.1 x 10-9)/(2.0 x 10-3)

[I-]2 = 3.55 x 10-6 [I-] = 1.88 x 10-3 mol dm-3

(1.88 x 10-3 mol dm-3) (10.0 dm3) = 1.88 x10-2 mol

(1.88 x10-2 mol) (166.00 g mol-1KI) = 3.12 g

How many grams of KI should be dissolved in 10.0 dm-3 of 0.00200 M Pb(NO3)2 solution so that precipitation of PbI2 (s) will just begin?

PbI2 Pb2+ (aq) + 2 I- (aq)

Ksp = [Pb2+] [I-]2 = 7.1 x 10-9

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Determining Whether Precipitation Occurs

–Example 2The concentration of calcium ion in blood plasma is 0.0025 M. If the concentration of oxalate ion is 1.0x10-7 M, do you expect calcium oxalate to precipitate? Ksp = 2.3x10-9.

CaC2O4 Ca2+ (aq) + C2O4

2- (aq)

[Ca2+ ] [C2O42- ] = 2.3x10-9

Three steps:

(1) Determine the initial concentrations of ions. (Put both in the Ksp formula)

(2) Evaluate the reaction quotient Qip.

(3) Compare Qip with Ksp. If Qip > Ksp then it will precipitate

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Determining Whether Precipitation Occurs – Another Example

In applying the precipitation criteria, the effect of dilution when solutions are mixed must be considered.

Example:

A 250.0 mL sample of 0.0012 M Pb(NO3)2 (aq) is mixed with 150.0 mL of 0.0640 M NaI (aq). Should precipitation of PbI2 (s), Ksp = 7.1x10-9, occur?

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Determining Whether Precipitation Is Complete A slightly soluble solid never totally precipitates from

solution, but we generally consider precipitation to be essentially complete if about 99.9% of the target ion is precipitated and only 0.1% or less is left in solution.

Three conditions that generally favor completeness of precipitation are:A very small value of Ksp.A high initial concentration of the target ion.A concentration of common ion that greatly

exceeds that of the target ion.

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Determining Whether Precipitation Is Complete - An Example

A 0.50 L solution of 0.0010 M BaCl2 is added to 0.50 L solution of 0.00010 M Na2SO4. Will the precipitation of SO4

2- as BaSO4 (s) be complete? Ksp = 1.1x10-10.

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Selective Precipitation

a) The first precipitate to form when AgNO3(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.

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Selective Precipitation An Example

Example 16.10An aqueous solution that is 2.00 M in AgNO3 is slowly added from a buret to an aqueous solution that is 0.0100 M in Cl- and also 0.0100 M in I-

AgCl (s) Ag+ (aq) + Cl- (aq) Ksp = 1.8x10-10

AgI (s) Ag+ (aq) + I- (aq) Ksp = 8.5x10-17

a. Which ion. Cl- or I-, is the first to precipitate from solution?

b. When the second ion begins to precipitate, what is the remaining concentration of the first ion?

c. Is separation of the two ions by selective precipitation feasible?

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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.

If the anion of the precipitate is that of a weak acid, the precipitate will dissolve somewhat when the pH is lowered; if, however, the anion of the precipitate is that of a strong acid, lowering the pH will have no effect on the precipitate.

Since CO2 is lost from the reaction of an insoluble carbonate with acid, this concept does not apply.

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Effect of pH on Solubility - An Example

Determine the molar solubility of Fe(OH)3 in

pH = 2.70 buffer. Ksp = 4.0x10-38 for Fe(OH)3.

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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.

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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: Pb2+, Hg2

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.

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Cation Group 1 (continued)Analyzing For Pb2+

Of the three possible ions in solution, PbCl2 is the most soluble in water.

The precipitate is washed with hot water and the washings then treated with aqueous K2CrO4.

If Pb2+ is present, chromate ion combines with lead ion to form a precipitate of yellow lead chromate, which is less soluble than PbCl2.

If Pb2+ is absent, then the washings just become tinged yellow but no precipitate is in evidence.

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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.

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Cation Group 1 (continued)Analyzing For Hg2

2+

When precipitate was treated with aqueous ammonia in the previous step, any Hg2

2+ underwent an oxidation-reduction reaction to form a dark gray mixture of elemental mercury and HgNH2Cl 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.

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Group 1 Cation Precipitates

left: cation goup 1 ppt: PbCl2, PbCl2, AgCl (all white)

middle: product from test for Hg2

2+: mix of Hg (black) and HgNH2Cl (white)

right: product from test for Pb2+: PbCrO4 (yellow) when K2CrO4(aq) is reacted with saturated PbCl2

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Hydrogen Sulfide In TheQualitative Analysis Scheme

H2S is a weak diprotic acid; there is very little ionization of the HS- ion and it is the precipitating agent.

H2S (aq) + H2O H3O+ (aq) + HS- (aq) Ka1 = 1.0x10-7

HS- (aq) + H2O H3O+ (aq) + S2- (aq) Ka2 = 1.0x10-19

H2S (aq) + H2O H3O+ (aq) + HS- (aq)

M2+ (aq) + HS- (aq) + H2O (l) MS (s) + H3O+ (aq)

Overall: M2+ (aq) + H2S (aq) + 2 H2O (l) MS (s) + 3 H3O+ (aq)

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Cation Groups 2, 3, 4 And 5

The concentration of HS- is so low in a strongly acidic solution, that only the most insoluble sulfides precipitate.

These include the eight metal sulfides of Group 2. Of the eight cations in Group 3, five form sulfides that

are soluble in acidic solution but insoluble in an alkaline ammonia/ammonium chloride buffer solution. The other three group 3 cation form hydroxide precipitates in the alkaline solution.

The cations of groups 4 and 5 form soluble sulfides, even in basic solution.

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Cation Groups 2, 3, 4 And 5

The hydroxides of Groups 4 and 5, with the exception of Mg2+, are also moderately or highly soluble.

The group 4 cations are precipitated as carbonates from a buffered alkaline solution.

The cations of group 5 remain soluble in the presence of all common reagents.

Within each of the qualitative analysis groups, additional reactions to dissolve group precipitates and to separate and selectively precipitate individual cations for identification and confirmation are used.

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Summary

The solubility product constant, Ksp, represents equilibrium between a slightly soluble ionic compound and its ions in a saturated aqueous solution.

The common ion effect is responsible for the reduction in solubility of a slightly soluble ionic compound.

Precipitation is assumed to be complete if no more than 0.1% of the target ion remains in solution.

The solubilities of some slightly soluble compounds depends strongly on pH.

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Summary

Precipitation, acid-base, and oxidation-reduction reactions, together with complex-ion formation, are all used extensively in the classical scheme for the qualitative analysis of common cations.

chapter sixteenprentice-hall ©2002slide 1 of 32 solubility products heterogeneous equilibria...

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