Chapter 18 Solubility and Complex-Ion Equilibria Dr. Peter Warburton [email protected] .

Chapter 18Solubility and Complex-Ion Equilibria

Dr. Peter [email protected]://www.chem.mun.ca/zcourses/1051.php

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Solubility equilibriaMmXx (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)

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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

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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.

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Problem

Write the expressions of Ksp of:a) AgCl

b) PbI2

c) Ca3(PO4)2

d) Cr(OH)3

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sp

234

32sp

22sp

sp

OHCrK

POCaK

IPbK

ClAgK

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Problem

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

10-

sp

25-sp

5-24

22sp

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2sp

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2 4

10 x 1.1 K

10 x 1.05 K

M 10 x 1.05 SO Ba x if xK

SOBaK

(aq)SO (aq) Ba (s)BaSO

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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.

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Molar solubility

Alternatively, if we know themolar solubility

OR the molar mass and the solubility

we can calculate

the Ksp for the solid.

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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.

(aq)Xx (aq)M m(aq)XM

(aq)XM (s)XMynondissociati

xm

xmOH

xm2

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Problem

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 149.885 gmol-1.

Answer: Ksp = 2 x 10-7

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Problem

A handbook lists the aqueous solubility of lithium phosphate (Li3PO4) as 0.034 g per 100 mL at 18 C. What is the Ksp of lithium phosphate at 18 C? The molar mass of Li3PO4 is 115.794 gmol-1.Answer: Ksp = 2.0 x 10-9

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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.

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Problem

How many milligrams of BaSO4 (molar mass is 233.391 gmol-1) are dissolved in a 225 mL sample of saturated aqueous barium sulphate? Ksp = 1.1 x 10-10 at 25 C.

Answer: mass = 0.55 mg

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The common-ion effect


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!

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Figure

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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!

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Mol

ar

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Problem

Calculate the molar solubility of MgF2 (Ksp = 7.4 x 10-11) in pure water and in 0.10 molL-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.

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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 10-20 M in the buffered solution.

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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!

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The diverse (“uncommon”) ion effect

The activities of ions tend to be LESS than the

concentration value as the total ionic

concentration increases.

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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”.

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Mol

ar

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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.

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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!

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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

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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.

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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.

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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.

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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!

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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

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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

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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!


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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!


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Mixing and equilibrium take time!

We must wait until dilution is completed

and equilibrium is established BEFORE

we say precipitation

occurred!

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Problem

Will a precipitate form when 0.150 L of 0.10 molL-1 Pb(NO3)2 and 0.100 L of 0.20 molL-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.

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Problem

How many drops (1 drop = 0.05 mL) of 0.20 M KI must we add to 100.0 mL of 0.010 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.

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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.

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Problem

A typical Ca2+ concentration in seawater is 0.010 M. Will the precipitation of Ca(OH)2 be complete from a seawater sample in which [OH-] is maintained at 0.040 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 0.010 M, the precipitation is not complete.

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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 1gL-1?

Molar mass Mg is 24.305 gmol-1

Ksp of Mg(OH)2 is 1.8 x 10-11

Answer: [OH-] needed is 1.6 x 10-2 M.

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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.

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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.

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In other words,

the concentration of Ag+

CAN NOT become large enough

to precipitate Ag2CrO4

until the AgBr

precipitation is complete.

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Problem

AgNO3 is slowly added to a solution with [Cl-] = 0.115 M and [Br-] = 0.264 M. What percent of the Br- remains unprecipitated at the point at which AgCl (s) begins to precipitate?

Ksp values

AgCl = 1.8 x 10-10 AgBr = 5.0 x 10-13

Answer: 0.12 % of Br- remains.

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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)

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Solubility and pH


- (aq) + H2O (l)

This reaction will shift to the right (products) as the pH becomes more acidic which means CO3

2- (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!

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Molar

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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


- (aq) + H2O (l)

K = Kb x 1/Kw = (2.1 x 10-4) x (1.0 x 1014)K = 2.1 x 1010

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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+!

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Problem

Will a precipitate of Fe(OH)3 form from a solution that is 0.013 M Fe3+ in a buffer solution that is 0.150 M acetic acid – 0.250 M acetate?

Ksp Fe(OH)3 = 4 x 10-38 Ka = 1.8 x 10-5

Answer: Since Qsp is 1 x 10-29, then precipitation will occur since Qsp>Ksp.

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Problem

What minimum [NH4+] must be present to

prevent precipitation of Mn(OH)2 (s) from a solution that is 0.0050 M MnCl2 and 0.025 M NH3? For Mn(OH)2 Ksp = 1.9 x 10-13 and Kb for NH3 is 1.8 x 10-5.

Answer: [NH4+] > 0.073 M

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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.

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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 10-10 = 3.1 x 10-3

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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).

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Molar

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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.

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Groups of precipitated ions

1) Chloride group – Pb2+, Ag+, Hg2

2+

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

+

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Reactions with hydrogen chloride

Most metal ions form soluble salts with chloride EXCEPT Pb2+, Hg2

2+, 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…

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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)

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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

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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) + CrO4

2- (aq)

PbCrO4(s)

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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

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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 behindMnS, FeS, Fe(OH)3, NiS, CoS, Al(OH)3,

ZnS, Cr(OH)3

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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

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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+

Chapter 18 Solubility and Complex-Ion Equilibria Dr. Peter Warburton [email protected] .

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Transcript of Chapter 18 Solubility and Complex-Ion Equilibria Dr. Peter Warburton [email protected] .