Solubility Equilibrium Chapter 7. The Solubility Equilibrium Remember from SPH3U: Solubility is the...

Solubility Equilibrium

Chapter 7

The Solubility Equilibrium

Remember from SPH3U: Solubility is the amount of solute that dissolves

in a given amount of solvent (usually water) at a particular temperature. Measured in g/100 mL or mol/L (molar concentration) Very soluble - > 10 g/100 mL

Soluble – 1 – 10 g/100 mLSlightly soluble – 0.1 – 1 g/100 mLInsoluble - < 0.1 g/100 mL


Produced solubility curvesMost solids increased in solubility as the

temperature increased. While, most gases decreased in solubility as the temperature increased.

Terminology – Saturated, Unsaturated, Supersaturated

The Solubility ProcessS

olu

bil

ity

(gm

/100

ml)

Temperature (oC)

Saturated solution

Unsaturated solution

Supersaturated solution


Dissociation of ionic compounds into ions Ionic equations, net ionic equations, spectator

ionsElectrolytes, non-electrolytes

Solubility chartsPredict precipitates in ionic solution

combinations (see periodic table given)

Solubility Product Constant – Ksp

The Ksp (solubility product constant) is the rate constant between a salt and its dissolved ions in a solution. The value indicated the degree to which a compound dissociated in water.

The higher the Ksp, the more soluble the compound.

The Solubility Product ConstantA solution equilibrium can be treated as a chemical equilibrium. The solubility equilibrium must be saturated and have

excess solute (salt), therefore they occur with low solubility compounds.

The balanced chemical reaction (dissociation equation) can be used to generate an equilibrium expression.

Remember, solids have a constant concentration and are not used in the equilibrium law expression. The equilibrium constant is simplified to the product of the ion concentration raised to their proper exponents.

The equilibrium constant (Keq) can be considered a solubility product constant (Ksp) in an solubility equilibrium.

The Solubility Product ConstantA solution equilibrium can be treated as a chemical equilibrium.

PbI2 (s) Pb 2+(aq) + 2 I 1-

(aq) (saturated sol’n)

PbI2 is a solid so its concentration does not alter and is therefore removed from the equilibrium expression.

PbI

IPb 1

2

2-112

eqK

2-12speq ][I ]Pb[K K

The Solubility Product Constant

A general application of the solubility equilibria results in the following Ksp expression.

BC(s) D b B+(aq) + c C-

(aq)

Where BC(s) is a slightly soluble salt, and B+(aq)

and C-(aq) are dissociated aqueous ions.

Ksp values are reported in reference tables such as those in the appendix. (Pg 802)

cb

sp CBK

The Solubility Product Constant usage

1. Calculating solubility constants from solubility data. (i.e.- solubility of 0.0003 g/100 mL @ 25oC)

2. Predicting whether a precipitate will form in a solubility equilibrium.

1. Trial ion product, Q– the reaction quotient applied to the ion concentrations.

2. Q > Ksp (supersaturated) precipitate will formQ = Ksp (saturated) precipitate will not formQ < Ksp (unsaturated) precipitate will not form

Calculating solubility constants from solubility dataDetermine the solubility of AgCl (s) at SATP given that its Ksp is 1.8X 10-10.

Strategy

1. Balanced chemical equation for the system.

2. Equilibrium Law expression.

3. Initial, Change in & Equilibrium Concentration table.

4. Sub-n-solve.


Strategy


AgCl(s) D Ag +1(aq) + Cl-1(aq)


2. Equilibrium Law expression.

Ksp = [Ag+1] [Cl-1] = 1.8 x 10-10


3. Initial, Change in & Equilibrium Concentration table.

AgCl (s) D Ag+1(aq) + Cl+1(aq)

[Initial] [constant] 0 0

[change in] [constant] +x +x

[Equilibrium] [constant] 0+x = x 0+x = x


4. Sub-n-solve.

Lmol 10 x 34.1

10 x 8.1

][10 x 8.1

]][[10 x 8.1

]Cl][Ag[

5-

10-

210-

10-

1(aq)

1(aq)

x

x

x

xx

K sp


4. Check assumptions

Lmol 10 x 34.1

][

]][[

]Cl][Ag[

5-

2

1(aq)

1(aq)

x

x

x

xx

K sp

The Solubility Product Constant usage

3. The common ion effect The reduction in the solubility of a salt

caused by the addition of a second salt with a common ion. Eg. The solubility of NaCl is decreased if

Cl- is already present in the solution Le Chatelier’s principle is applied and

results in a shift of the solubility equilibrium.

Use equilibrium (ICE) table to solve for new concentrations.

ExampleCalculate the molar solubility of PbSO4 in 0.100 M Na2SO4 at SATP. Ksp = 1.96 x 10-8

Split this problem into two steps. Initially, NaSO4(aq) has already dissociated in water to a concentration of 0.100 mol/L

Na2SO4 (s) 2Na+(aq) + SO4

2- (aq)

The [SO42-

(aq)] = [Na2SO4 (s)] = 0.100M

PbSO4 is then added to this solution.

PbSO4 D Pb2+(aq) + SO4

2- (aq)

Because there are already SO42- ions present in the

solution, the PbSO4 cannot completely dissociate in this solution. We cannot use the assumption that the concentration of ions will be equal to the concentration of PbSO4.

Let molar solubility of PbSO4 be x.

Ksp = [Pb2+] [SO42-] = 1.96´ 10-8

PbSO4(s) ® Pb2+(aq) + SO4

2-(aq)

Initial -- 0 0.100

Change-- x + x

Equil. -- x 0.100 + x

Ksp = [Pb2+] [SO42-]

(x)(0.100 + x) = 1.96´ 108

x2 + 0.100 x -1.96´ 108 = 0

Using the quadratic equation, x = 2.00 x 10-7 M

Molar solubility of PbSO4 in 0.100 M Na2SO4

= 2.0 ´ 107 M

Molar solubility of PbSO4 in pure water = 1.40´104 M

Note the solubility is very much reduced in 0.100 M Na2SO4because of the common ion effect.

Predicting Precipitate

Reaction Quotient (Q):Used to determine if a solubility reaction has

reached saturation. It tells you how far a reaction has proceeded.

Using Q

To do so, the concentration of ions in solution can be used to calculate a trial ion product and then compare it to the listed value of Ksp Given a reaction AB(s) D A+

(aq) + bB-(aq)

Trial Ion Product = [A+][B-]b

Determining if a precipitate will formStrategy


2. Identify the concentrations to be considered.

3. Use the equilibrium law expression to generate the “Trial Ion Product (Qsp)” expression.

4. Use given concentration and the expression to solve for Qsp.

Determining if a precipitate will formStrategy

5. Compare the Qsp value to the known Ksp value to determine if the situation is at equilibrium. Qsp = Ksp the solution is saturated & at equilibrium

so no precipitate will form. Qsp < Ksp the solution is unsaturated so no

precipitate will form. Qsp > Ksp the solution is “supersaturated” so a

precipitate will form.

Comparing Q to Ksp

Determining if a precipitate will formIf 50 mL of 0.200 mol/L NiCl2 (aq) and 50 mL of 0.0500 mol/L Na2CO3 (aq) combined at 25oC will a precipitate form? Ksp for NiCO3 (s) is 8.2 x 10-4.

NiCl2 (aq) + Na2CO3 (aq) → NiCO3 (s) + 2 NaCl (aq)

(Use the solubility rules to identify which of the products is most likely to form a solid.)

NiCO3 (s) ⇌ Ni 2+(aq) + 2 Cl-(aq)


NiCl2 (aq) → Ni 2+(aq) + 2 Cl- (aq)

[Ni2+]= 0.200 mol/L(50 mL)/(100 mL) = 0.100 mol/L

Na2CO3 (aq) → 2 Na+(aq) + CO3

2- (aq)

[CO32-]= 0.0500 mol/L(50 mL)/(100 mL) = 0.0250 mol/L


NiCO3 (s) ⇌ Ni 2+(aq) + CO3

2- (aq)

0025.0

)mol0250.0)(Lmol 100.0(

]][CONi[ -23

2

L

Qsp


Qsp > Ksp (0.0025 > 8.2 x 10-4 )

In this case the solution is either supersaturated or a precipitate will be formed.

Solubility Equilibrium Chapter 7. The Solubility Equilibrium Remember from SPH3U: Solubility is the...

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