 # Chapter 18 Solubility and Complex-Ion Equilibria

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Chapter 18 Solubility and Complex-Ion Equilibria
Dr. Peter Warburton

Solubility equilibria
MmXx (s)  m Mn+ (aq) + x Xy- (aq) The equilibrium is established when we have a saturated solution of ions forming the solid and solid is dissociating to form the ions in solution. The rates of these processes must be equal. (eq’m definition)

Solubility equilibria
For a dissolution process, we give the equilibrium constant expression the name solubility product (constant) Ksp. For MmXx (s)  m Mn+ (aq) + x Xy- (aq) Ksp = [Mn+]m [Xy-]x

Ksp is an equilibrium constant
Since Ksp is an equilibrium constant we MUST refer to a specific balanced equation (by definition this balanced equation is one mole of solid becoming aqueous ions) at a specific temperature.

Problem Write the expressions of Ksp of: a) AgCl b) PbI2 c) Ca3(PO4)2
d) Cr(OH)3

Problem If a saturated solution of BaSO4 is prepared by dissolving solid BaSO4 in water, and [Ba2+] = 1.05 x 10-5 molL-1, what is the Ksp for BaSO4?

Molar solubility If we know the Ksp value for a solid, we can calculate the molar solubility, which is the number of moles of the solid that can dissolve in a given amount of solvent before the solution becomes saturated. The molar solubility leads to the solubility (by using the molar mass) which is the mass of the solid that can dissolve in a given amount of solvent before the solution becomes saturated.

Alternatively, if we know the molar solubility OR
the molar mass and the solubility we can calculate the Ksp for the solid.

Molar solubility A good (but NOT physically correct) way to think about molar solubility is to treat our solid dissolution AS IF there are two separate processes: The molar solubility has the same value as the concentration of aqueous MmXx after the first step. We can figure out this concentration based on concentrations of ions in the second step.

Problem Answer: Ksp = 2 x 10-7
A handbook lists the aqueous solubility of AgOCN as 7 mg per 100 mL at 20 C. What is the Ksp of AgOCN at 20 C? The molar mass of AgOCN is gmol-1. Answer: Ksp = 2 x 10-7

Problem Answer: Ksp = 2.0 x 10-9
A handbook lists the aqueous solubility of lithium phosphate (Li3PO4) as g per 100 mL at 18 C. What is the Ksp of lithium phosphate at 18 C? The molar mass of Li3PO4 is gmol-1. Answer: Ksp = 2.0 x 10-9

Problem Which has the greater molar solubility:
AgCl with Ksp = 1.8 x 10-10 or Ag2CrO4 with Ksp = 1.1 x 10-12? Answer: The molar solubility of AgCl is 1.3 x 10-5 M while the molar solubility of Ag2CrO4 is 6.5 x 10-5 M. Silver chromate has a higher molar solubility.

Problem How many milligrams of BaSO4 (molar mass is gmol-1) are dissolved in a 225 mL sample of saturated aqueous barium sulphate? Ksp = 1.1 x at 25 C. Answer: mass = 0.55 mg

MmXx (s)  m Mn+ (aq) + x Xy- (aq)
The common-ion effect MmXx (s)  m Mn+ (aq) + x Xy- (aq) If we have dissolved a solid in pure water and we add to this solution another solution containing one of the common ions, then Le Chatalier’s Principle tells us what will happen: The presence of the common-ion in the added solution will force the dissolution reaction to the left, meaning more solid will form!

The common-ion effect

Figure

MmXx (s)  m Mn+ (aq) + x Xy- (aq)
The common-ion effect MmXx (s)  m Mn+ (aq) + x Xy- (aq) If instead of dissolving a solid in pure water we try and dissolve it into a solution that already contains one of the common ions, then Le Chatalier’s Principle tells us what will happen: The presence of the common-ion already in solution will force the dissolution reaction to the left, meaning less solid will dissolve than would dissolve in pure water!

Molar

Problem Calculate the molar solubility of MgF2 (Ksp = 7.4 x 10-11) in pure water and in 0.10 molL-1 MgCl2 at 25 °C. Answer: The molar solubility is 2.6 x 10-4 M in pure water and 1.4 x 10-5 M in 0.10 M magnesium chloride.

Problem What is the the molar solubility of Fe(OH)3 (Ksp = 4 x 10-38) in a buffered solution with pH = 8.20 at 25 °C. Answer: The molar solubility is 1 x M in the buffered solution.

Limitations of Ksp If our solid is more than slightly soluble then we really should use the activities of our ions in solution rather than concentrations. These two measures are nearly the same for very dilute ion concentrations, but can become quite different at higher concentrations!

The diverse (“uncommon”) ion effect
The activities of ions tend to be LESS than the concentration value as the total ionic concentration increases.

The diverse (“uncommon”) ion effect
Adding a salt that does NOT feature a common ion to the solution will tend to decrease the activity (the “effective” concentration) of the ions in solution. The ion concentrations appear smaller than they should be at equilibrium so more solid dissolves to reach the appropriate equilibrium “concentrations”.

Molar

Ion-pair formation and Ksp
We assume in Ksp calculations that the solid dissociates completely into ions in solution. If this is not true, then the ionic concentrations we measure do not include dissolved but undissociated molecules or ion pairs which come from solid dissolution.

Ion-pair formation and Ksp
Positive ions and negative ions are attracted to each other and so they can form an ion pair that has a chemical identity different from each of the individual ions!

Ion-pair formation and Ksp
Say we measure a molar solubility for magnesium fluoride to be 4 x 10-3 M and assume that [Mg2+] = 4 x 10-3 M and [F-] = 8 x 10-3 M to give a calculated Ksp of 3 x 10-7

Ion-pair formation and Ksp
IN REALITY the presence of undissociated MgF2 (aq) and MgF- ion pairs means that not all of the solid that has dissolved is found as free ions, so our ionic concentrations are LOWER than we assumed, and so Ksp is actually smaller than we calculated.

Simultaneous equilibria
We’ve seen in reference to Le Chatalier’s Principle that if more than one reaction can take place in a container, then the reactions might not be able to be treated independently. Other equilibrium processes may affect the solubility of the solid and lead to miscalculated Ksp values.

Simultaneous equilibria
For example, we’ve seen that AgI becomes more soluble when we add ammonia because of the formation of a complex of silver ions and ammonia. We’ll look more closely at complex formation a little later.

Assessing the limitations of Ksp
Most tabulated Ksp values are actually based on activities, and not concentrations. We use concentrations in our examples, so our calculations represent an ideal, and not reality. Generally we could be in error by over 100 times!

Criteria for precipitation and its completeness
Can we predict if a solid will form if we mix two solutions of different ions? Consider the mixing of two different solutions, one with Ca2+ ions and one with F- ions. A formation of solid is the dissolution reaction in reverse, so we can express the reaction using the dissolution equation CaF2 (s)  Ca2+ (aq) + 2 F- (aq) Ksp = [Ca2+ ][F-]2

Criteria for precipitation and its completeness
When we mix the solutions (BE CAREFUL – mixing ALWAYS changes the concentrations of both our ions!) the system is most likely not at equilibrium. Like in other equilibrum problems, we can use a reaction quotient Qsp (often called the ion product) to tell us in which direction the system must go to reach equilibrium Qsp = [Ca2+ ][F-]2

No precipitation occurs!
Criteria for precipitation and its completeness If Qsp > Ksp, the solution is supersaturated, so the system is not at equilibrium. The concentration of the ions is greater than it would be at equilibrium, and so the reaction wants to shift from ions towards the solid. We expect precipitation to occur! If Qsp = Ksp, the solution is saturated, and the system is at equilibrium. No precipitation occurs!

No precipitation can occur!
Criteria for precipitation and its completeness If Qsp < Ksp, the solution is unsaturated, so the system is not at equilibrium. The concentration of the ions is less than it would be at equilibrium, and so the reaction wants to shift from solid towards the ions. No precipitation can occur!

Mixing and equilibrium take time!
We must wait until dilution is completed and equilibrium is established BEFORE we say precipitation occurred!

Problem Will a precipitate form when L of 0.10 molL-1 Pb(NO3)2 and L of 0.20 molL-1 NaCl are mixed? Ksp of PbCl2 is 1.2 x 10-5 Answer: Qsp = 3.8 x 10-4 > Ksp so precipitation should occur.

Problem How many drops (1 drop = 0.05 mL) of 0.20 M KI must we add to mL of M Pb(NO3)2 to get precipitation of lead iodide to start? Ksp of PbI2 is 7.1 x 10-9 Answer: We require at least 9 drops.

Complete precipitation
Generally we treat precipitation as complete if 99.9% of the original ion concentration has been lost to the precipitate. For example, if our initial [Pb2+] is 0.10 M, then precipitation by adding I- is complete when our solution contains a [Pb2+] less than 1 x 10-4 M.

Problem A typical Ca2+ concentration in seawater is M. Will the precipitation of Ca(OH)2 be complete from a seawater sample in which [OH-] is maintained at M? Ksp of Ca(OH)2 is 5.5 x 10-6 Answer: Since the final [Ca2+] is 3.4 x 10-3 M, which is 34 % of M, the precipitation is not complete.

Problem What [OH-] should be maintained in a solution if, after precipitation of Mg2+ as solid magnesium hydroxide, the remaining [Mg2+] is to be at a level of 1gL-1? Molar mass Mg is gmol-1 Ksp of Mg(OH)2 is 1.8 x 10-11 Answer: [OH-] needed is 1.6 x 10-2 M.

Fractional precipitation
If we have a solution with both CrO42- ions and Br- ions and add a large amount of Ag+ ions at once, then both Ag2CrO4 and AgBr will precipitate in our container at the same time.

Fractional precipitation
If we slowly add the Ag+ solution instead the solid with the significantly lower molar solubility (AgBr in this case – do the calculations to check this for yourself) will precipitate first and consume the added Ag+ preferentially.

Fractional precipitation
In other words, the concentration of Ag+ CAN NOT become large enough to precipitate Ag2CrO4 until the AgBr precipitation is complete.

Fractional precipitation

Problem AgNO3 is slowly added to a solution with [Cl-] = M and [Br-] = M. What percent of the Br- remains unprecipitated at the point at which AgCl (s) begins to precipitate? Ksp values AgCl = 1.8 x AgBr = 5.0 x 10-13 Answer: % of Br- remains.

Solubility and pH If a solid dissolves to give a basic anion in solution, addition of strong acid will increase the solubility of the solid. CaCO3 (s)  Ca2+ (aq) + CO32- (aq) Ksp = 2.8 x 10-9 Carbonate, CO32-, is a basic anion that will react with a proton to give HCO3- CO32- (aq) + H3O+ (aq)  HCO3- (aq) + H2O (l)

Solubility and pH More solid will dissolve!
CO32- (aq) + H3O+ (aq)  HCO3- (aq) + H2O (l) This reaction will shift to the right (products) as the pH becomes more acidic which means CO32- (aq) decreases. However, in our first equilibrium IT ALSO MUST DECREASE so the first equilibrium will also shift to the right to compensate. More solid will dissolve!

Molar

CaCO3 (s)  Ca2+ (aq) + CO32- (aq)
Adding equilibria A better way to state the effect of pH on solubility comes when we add dissolution and weak base – strong acid reactions together: CaCO3 (s)  Ca2+ (aq) + CO32- (aq) Ksp = 2.8 x 10-9 CO32- (aq) + H3O+ (aq)  HCO3- (aq) + H2O (l) K = Kb x 1/Kw = (2.1 x 10-4) x (1.0 x 1014) K = 2.1 x 1010

The solubility of the solid will increase in the presence of H3O+!
Adding equilibria CaCO3 (s) + H3O+ (aq)  Ca2+ (aq) + HCO3- (aq) + H2O (l) K’ = K x Ksp = (2.1 x 1010) x (2.8 x 10-9) K’ = 59 The solubility of the solid will increase in the presence of H3O+!

Problem Will a precipitate of Fe(OH)3 form from a solution that is M Fe3+ in a buffer solution that is M acetic acid – M acetate? Ksp Fe(OH)3 = 4 x Ka = 1.8 x 10-5 Answer: Since Qsp is 1 x 10-29, then precipitation will occur since Qsp>Ksp.

Problem Answer: [NH4+] > 0.073 M
What minimum [NH4+] must be present to prevent precipitation of Mn(OH)2 (s) from a solution that is M MnCl2 and M NH3? For Mn(OH)2 Ksp = 1.9 x and Kb for NH3 is 1.8 x 10-5. Answer: [NH4+] > M

Formation of complex ions
Solubility of a solid increases if there is the ability to form a complex ion. An example of a complex ion is Ag(NH3)2+. Such complexes affect solubility by reducing the concentration of the cation so that the dissolution reaction must shift to the products to replace the cation concentration to re-establish equilibrium.

Formation of complex ions
Ag+ (aq) + 2 NH3 (aq)  Ag(NH3)2+ (aq) Kf = 1.7 x 107 AgCl (s)  Ag+ (aq) + Cl- (aq) Ksp = 1.8 x 10-10 In the presence of ammonia, the dissolution of AgCl can be expressed by the sum of these two reactions 2 NH3 (aq) + AgCl (s)  Ag(NH3)2+ (aq) + Cl- (aq) K = Kf x Ksp = 1.7 x 107 x 1.8 x = 3.1 x 10-3

Formation of complex ions
We see the dissolution of AgCl occurs to a greater level of completion in the presence of ammonia (K = 3.1 x 10-3) than it does in pure water (Ksp = 1.8 x 10-10).

Molar

Qualitative cation analysis
Qualitative analysis is concerned with “what do we have?” and NOT “how much do we have?” If we want to identify what cations we have in a solution, we can use a series of precipitation reactions in a certain order to tell us.

Groups of precipitated ions
1) Chloride group – Pb2+, Ag+, Hg22+ 2) Hydrogen Sulphide group- Pb2+, Hg2+, Bi3+, Cu2+, Cd2+, As3+, Sn2+, Sb3+ 3) Ammonium sulphide group – Mn2+, Fe2+, Fe3+, Ni2+, Co2+, Al3+, Zn2+, Cr3+ 4) Carbonate group – Mg2+, Ca2+, Sr2+, Ba2+ 5) Soluble group – Na+, K+, NH4+

Reactions with hydrogen chloride
Most metal ions form soluble salts with chloride EXCEPT Pb2+, Hg22+, and Ag+. Adding aqueous HCl to our unknown solution will let us now if we have one or more of these ions because we will get white precipitate(s). If we want to know if we have more than one of these ions, we do further tests on the precipitated solids…

Further tests for insoluble chlorides
In a) we have a mixture of AgCl, Hg2Cl2, and PbCl2. If we add ammonia, any AgCl should dissolve because of complex ion formation AgCl (s) + 2 NH3 (aq)  [Ag(NH3)2]+ (aq) + Cl- (aq)

Further tests for insoluble chlorides
Also after adding ammonia (b), any Hg2Cl2 will give us a grey solid that is a mixture of black liquid Hg and white solid HgNH2Cl: Hg2Cl2 (s) + 2 NH3 (aq)  Hg (l) + HgNH2Cl (s) + NH4Cl (aq) black white

Further tests for insoluble chlorides
Adding chromate (CrO42-) (c) to a Pb2+ solution derived by heating the precipitate solutions (it’s the most soluble) will give a yellow precipitate - PbCrO4 Pb2+ (aq) + CrO42- (aq)  PbCrO4(s)

Reactions with hydrogen sulfide
S2- is capable of giving precipitates of many ions. H2S is a potential source of S2- in solution because it is a diprotic acid H2S (aq) + H2O (l)  H3O+ (aq) + HS- (aq) Ka1 = 1.0 x 10-7 HS- (aq) + H2O (l)  H3O+ (aq) + S2- (aq) Ka2 = 1 x 10-19

Reactions with hydrogen sulfide
However, in acidic solution (with HCl), some of the precipitates dissolve, leaving behind PbS, HgS, Bi2S3, CuS, CdS, As2S3, SnS, Sb2S3 In basic solution (by adding ammonia) these precipitates dissolve, leaving behind MnS, FeS, Fe(OH)3, NiS, CoS, Al(OH)3, ZnS, Cr(OH)3

Reactions with carbonate
Addition of carbonate ion (CO32-) in basic solution (usually with an ammonia-ammonium buffer) will precipitate the alkali earth metal carbonates (see Chapter 21) CaCO3, MgCO3, SrCO3, BaCO3

Na+, K+, and NH4+ The soluble group
Any ions left in solution after the first four reaction groups are tested for are the cations of soluble salts Na+, K+, and NH4+