# Chapter 7  Solubility

## 7.1  Solubility

Solubility

• qualitative definition = whether a solute dissolves in a solvent (soluble) or does not (insoluble)
• quantitative definition = the amount of a solute that will dissolve in a given amount of solvent (e.g. 359 g NaCl / 1.00 L H2O)

Dissolution = the reaction of dissolving

• covalent molecules, e.g. C6H12O6 (s) ⇋ C6H12O6 (aq)
• ionic compounds, e.g. NaCl (s) ⇋ Na+ (aq) + Cl (aq)

Precipitation = the formation of a solid from the dissolved solute, the reverse reaction of dissolution

## 7.2  Solubility Product Constant and Molar Solubility

The Solubility Product Constant, Ksp, for ionic compounds, MxAy (s) ⇋ x Mm+ (aq) + y Aa (aq) is:

 Ksp = [Mm+]x[Aa−]y     (1)

Ksp is related to the Molar Solubility, s, which is the molarity of MxAy in a saturated solution of MxAy.
What is the Molar Solubility of NaCl? What is the molarity of a saturated solution of NaCl?

The solubility of NaCl is 361.24 g NaCl/1.00 L H2O with a solution density of 1.1978 g/mL.

1000 mL H2O ×
 0.99705 g H2O 1 mL H2O
= 997.05 g H2O

 997.05 g H2O + 361.24 g NaCl = 1358.29 g solution
Vsolution =
1358.29 g solution ×
 1 mL 1.1978 g
×
 1 L 1000 mL
= 1.1340 L

Which allows the calculation of the Molar Solubility, s, which is the concentration of NaCl in a saturated solution of NaCl:

s =
[NaCl] =
361.24 g NaCl×
 mol NaCl 58.4425 g NaCl
1.1340 L solution
= 5.451 M NaCl

Using reaction stoichiometry we can get the individual concentrations of the dissolved ions:

5.451 M  NaCl ×
 1 Na+ 1 NaCl
= 5.451 M Na+

5.451 M NaCl ×
 1 Cl− 1 NaCl
= 5.451 M Cl

which allow you to calculate the Ksp = [Na+][Cl]:

 Ksp = (5.451 M Na+)(5.451 M Cl−) = 29.71

NaCl is often considered soluble in water, and here we see it has an approximate Ksp = 29.71, which is greater than 1 indicating a product favored reaction which is consistent with it being considered soluble.
There are many compounds that are considered insoluble, with Ksp < 1. For example, what are the equilibrium concentrations of Pb2+ and Br in a saturated solution of PbBr2 with Ksp = 4.67 × 10−6? What is the molar solubility of PbBr2?

 R PbBr2 (s) ⇋ Pb2+ (aq) + 2 Br− (aq) Ksp = [Pb2+][Br−]2 I 0.0 M 0.0 M C +x +2x E x 2x Ksp = (x)(2x)2 = (x)(4x2) = 4x3 = 4.67 × 10−6
x =
 ∛
 4.67 × 10−6 4
=

 4.67 × 10−6 4

 1/3
= 0.01053
 [Pb2+] = x = 0.01053  M Pb2+
 [Br−] = 2x = 2(0.01053) = 0.02106  M Br−

What is the molar solubility, s, of PbBr2?

 s = x = 0.01053  M PbBr2

Given the above example, it is very easy to figure out what the result of the RICE table will be. Notice how very different solubility product constants, Ksp, result in very similar molar solubilities, s. Some examples:

 AgI Ksp = [Ag+][I−] Ksp = (x)(x) = x2 = 8.51 × 10−17 s = x = 9.22 × 10−9 PbI2 Ksp = [Pb2+][I−]2 Ksp = (x)(2x)2 = 4x3 = 9.8 × 10−9 s = x = 1.3 × 10−3 Al(OH)3 Ksp = [Al3+][OH−]3 Ksp = (x)(3x)3 = 27x4 = 1.3 × 10−33 s = x = 2.6 × 10−9 Ba3(PO4)2 Ksp = [Ba2+]3[PO43−]2 Ksp = (3x)3(2x)2 = 108x5 = 6 × 10−39 s = x = 9 × 10−9

When comparing solubilities between compounds, you should compare molar solubilities, s. You cannot use Ksp to compare solubilities between two compounds unless they have the same reaction stoichiometry.

## 7.3  Common Ion Effect

The common ion effect is the influence a dissolved solute will have on the solubility of another solute when they share a common ion in their chemical formulas.
The molar solubility of PbBr2 calculated earlier was s = 0.01053 M PbBr2. This is true if PbBr2 is being dissolved in pure water.
What is the solubility of PbBr2 in a 0.200 M NaBr solution, which contains a common ion, Br?

 R PbBr2 (s) ⇋ Pb2+ (aq) + 2 Br− (aq) Ksp = [Pb2+][Br−]2 I 0.0 M 0.200 M C +x +2x E x 0.200 + 2x Ksp = (x)(0.200 + 2x)2 = 4.67 × 10−6

This is a cubic equation that could be complicated to solve but we understand that since Ksp is so small, the distance to equilibrium is small, and so x is small (x ≈ 0):

 0.200 + 2x ≈ 0.200
 (x)(0.200)2 = 4.67 × 10−6
x =
 4.67 × 10−6 (0.200)2
= 1.17 × 10−4

We can see this is a great approximation (0.200 + 2x ≈ 0.200) because if we iterate:

(x)
0.200 + 2
1.17 × 10−4

 2
= 4.67 × 10−6
x =
4.67 × 10−6

0.200 + 2
1.17 × 10−4

 2
= 1.16 × 10−4

We see the answer has effectively converged to x = 1.16 × 10−4. The molar solubility of PbBr2 in 0.200 M NaBr is less than in pure water:

 s = x = 1.16 × 10−4 M PbBr2 in 0.200 M NaBr < 0.01053 M PbBr2 in pure water

The common ion effect here is that the NaBr suppressed the solubility of PbBr2 due to the common ion, Br.

## 7.4  Acid/Base Considerations

Many ions exhibit acid/base behavior when dissolved in water. If a compound is being dissolved in an acidic or basic solution, it has the capacity to react with the higher H3O+ or OH concentrations present in the solution.
Some metal hydroxides (Fe(OH)2, Mg(OH)2, Ca(OH)2) will naturally have lower solubility in basic solutions due to the common ion effect, the common ion being OH.
Some metal hydroxides (Zn(OH)2, Pb(OH)2, Sn(OH)2, Co(OH)2, Al(OH)3) have higher solubility in basic solutions due to the formation of a complex anion which consumes the common ion, OH, and the metal ion, leading to an increase in solubility since the complex anion does not appear in the expression for Ksp.

 Zn(OH)2 (s) ⇋ Zn2+ (aq) + 2 OH− (aq) + Zn2+ (aq) + 4 OH− (aq) ⇋ [Zn(OH)4]2− (aq) Zn(OH)2 (s) + 2 OH− (aq) ⇋ [Zn(OH)4]2− (aq)

Metal hydroxides and compounds with strong or weak conjugate base anions will have higher solubility in acidic solutions because the OH or A will continually be consumed by the high levels of H3O+ via acid-base neutralization, effectively allowing more "room" for the compound to dissolve.

 Zn(OH)2 (s) ⇋ Zn2+ (aq) + 2 OH− (aq) + 2 H3O+ (aq) + 2 OH− (aq) ⇋ 4 H2O (l) Zn(OH)2 (s) + 2 H3O+ (aq) ⇋ Zn2+ (aq) + 4 H2O (l)

 FeS (s) ⇋ Fe2+ (aq) + S2− (aq) + H3O+ (aq) + S2− (aq) ⇋ HS− (aq) + H2O (l) FeS (s) + H3O+ (aq) ⇋ Fe2+ (aq) + HS− (aq) + H2O (l)

## 7.5  Complex Ion Formation

The formation of complex ions can lead to increased solubility for insoluble compounds.

KspAgCl = 1.77 × 10−10 which leads to s = 1.33 × 10−5 M AgCl in pure water.

If we were to try dissolve it in a solution of ammonia, NH3 (0.100 M), we would see the formation of a complex ion which would increase the apparent solubility of AgCl by consuming the Ag+ ions:

 Ag+ (aq) + 2 NH3 (aq) ⇋ [Ag(NH3)2]+ (aq) Kf = 1.7 × 107 + AgCl (s) ⇋ Ag+ (aq) + Cl− (aq) Ksp = 1.77 × 10−10 R AgCl (s) + 2 NH3 (aq) ⇋ [Ag(NH3)2]+ (aq) + Cl− (aq) Kf × Ksp = 3.01 × 10−3 I 0.100 0.0 0.0 M C −2x +x +x E 0.100 − 2x x x
K =

 ⎡ ⎣ ⎡ ⎣ Ag(NH3)2 ⎤ ⎦ + ⎤ ⎦ [Cl−]
[NH3]2

3.01 × 10−3 =
 (x)(x) (0.100 − 2x)2

 s = x = 4.94 × 10−3

The molar solubility of AgCl in 0.100 M NH3 is 4.94 × 10−3/1.33 × 10−5 = 371 times higher than in pure water.

## 7.6  Precipitation and Qsp

Qsp indicates whether a solution is saturated or not

• Qsp < Ksp = unsaturated solution; no precipitation
• Qsp = Ksp = saturated solution; no precipitation
• Qsp > Ksp = supersaturated solution; precipitation is likely
##### Problem

If 22.0 mL 3.02 × 10−4 M FeF3 is combined with 15.0 mL 7.96 × 10−4 M KOH, is there a precipitate formed?
FeF3 (aq) + 3 KOH (aq) ⇋ Fe(OH)3 (s) + 3 KF (aq)

Need to calculate initial concentrations of the ions of the potential precipitate, Fe(OH)3, in the new total volume to calculate Qsp=[Fe3+]0[OH]03.

 15.0 mL + 22.0 mL = 37.0 mL
[Fe3+]0 = 3.02 × 10−4 M FeF3 ×
 1 Fe3+ 1 FeF3
×
 22.0 mL 37.0 mL
= 1.80 × 10−4 M Fe3+

[OH]0 = 7.96 × 10−4 M KOH ×
 1 OH− 1 KOH
×
 15.0 mL 37.0 mL
= 3.23 × 10−4 M OH

Precipitation is the reverse of dissolution, so we’ll use the dissolution reaction with its Ksp:

 R Fe(OH)3 (s) ⇋ Fe3+ (aq) + 3 OH− (aq) Ksp = [Fe3+][OH−]3 = 2.79 × 10−39 I 1.80 × 10−4 3.23 × 10−4 Qsp = (1.80 × 10−4)(3.23 × 10−4)3 = 6.07 × 10−15 C −x −3x E 1.80 × 10−4 - x 3.23 × 10−4 - 3x Ksp = (1.80 × 10−4 − x)(3.23 × 10−4 − 3x)3

We see that Qsp = 6.07 × 10−15 > Ksp = 2.79 × 10−39 and so the solution is supersaturated and therefore not at equilibrium. Precipitation will occur as the solution moves towards equilibrium and x is the amount of Fe(OH)3 that is going to precipitate, expressed as a concentration (i.e. Δ [Fe(OH)3]precipitates).

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### 7.6.1  Sequential Precipitation

##### Problem

A solution is 0.100 M KI and 0.200 M KBr. If Ag+ is added to the solution (from AgNO3), which precipitates first, AgI (Ksp = 8.51 × 10−17) or AgBr (Ksp = 5.35 × 10−13)?

AgI and AgBr have the same dissolution stoichiometry so we can directly compare their Ksp without calculating their molar solubilities, s. AgI is less soluble than AgBr and since the concentrations of the I and Br are so similar we can expect AgI to precipitate first.

———————————————————————–

##### Problem

Above what concentration of Ag+ will AgI start to precipitate?

 Ksp =  [Ag+][I−] = [Ag+](0.100 M I−) = 8.51 × 10−17
[Ag+] =
 8.51 × 10−17 0.100
= 8.51 × 10−16 M Ag+
needed for AgI to precipitate

———————————————————————–

##### Problem

Above what concentration of Ag+ will AgBr start to precipitate?

 Ksp = [Ag+][Br−] = [Ag+](0.200 M Br−) = 5.35 × 10−13
[Ag+] =
 5.35 × 10−13 0.200
= 2.68 × 10−12 M Ag+
needed for AgBr to precipitate

We can see that it takes a much lower [Ag+] to initiate precipitation of AgI so it will precipitate first.

———————————————————————–

##### Problem

What is the [I] at the point that AgBr starts to precipitate?

When Ag+ is added to this solution, AgI will precipitate first eventually followed by AgBr once the [Ag+] gets high enough for AgBr to precipitate. In order to get a large [Ag+], a lot of the I needs to precipitate first, leaving a low [I]. The product of [Ag+] and [I] must equal Ksp = [Ag+][I], so you can solve for the [I] that would be present alongside the [Ag+] needed to start precipitating AgBr by:

 8.51 × 10−17 = (2.68 × 10−12 M Ag+)[I−]
[I] =
 8.51 × 10−17 2.68 × 10−12 M Ag+
= 3.18 × 10−5 M I when AgBr starts precipitating

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