Solubility
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Readings for this section

Introduction 09/02/2007

We've seen many ionic compounds. Many are soluble in water and many are not. There is no simple set of rules which we can use to predict which ones will be soluble or not.

Some useful trends have been observed.

• multiple-charged ions are less soluble than single-charged ions.
• smaller ions are more soluble than larger ions.

For example,

• cations like NH₄⁺ and all alkali metal ions are small, singly charged and are all soluble.
• anions like Cl^–, Br^–, I^–, NO₃^– and ClO₄^– are all soluble except with Mg²⁺ and Ag⁺.
• Most hydroxides are insoluble (except with alkali metals as in NaOH or KOH).
• Salts of doubly charged ions CO₃^2–, S^2–, PO₄^3– mostly are insoluble.
• Salts of singly charged versions of these are soluble (HCO₃^–, H₂PO₄^– )

Reactions where soluble compounds react to form insoluble ones are called precipitation reactions. The reverse is a dissolution reaction where solid compounds dissolve upon addition into water.

For a table of solubilities click here

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Solubility Product Constant

Since salts dissolve into electrically charged species (as do acids and bases) they fall in the general category called electrolytes.  We saw in the case of acids and bases that some dissociate completely into ions and are called strong while others dissociate only partially and are called weak. This terminology applies to salts as well. A strong electrolyte is any electrolyte (acid, base, salt) that dissociates 100% into ions. A weak electrolyte sets up an equilibrium in water so that some exits as the original undissociated species and some exists as ions in solution or as undissolved solid.

In reactions involving ions which have the potential to form insoluble compounds, we must consider equilibrium conditions in order to decide whether or not a precipitate (ppt) will form.

for example, the reaction between the calcium cation and the sulphate anion is written as follows (always put the solid as the reactant and the ions as the product.)

CaSO4(s)  Ca2+(aq) + SO42–(aq)

Recall that the CaSO₄ is a pure solid and  therefore the activity is 1. Since this equilibrium constant is the product of the ion concentrations (solubility products) we use subscript sp for this new value.

K_sp = [Ca²⁺][SO₄^2–]

K_sp is the solubility product constant and the right hand side of this equation is often called the ion product.

In general, for the reaction

A_xB_y(s) X A^m+(aq) + Y B^n–(aq)

we have

K_sp = [A^m+]^X[B^n–]^Y @ equilibrium.

We can perform equilibrium type calculations as we did previously.

Example:

CaSO₄ at 25ºC in water has a solubility of 4.9×10^–3 M, i.e., enough CaSO₄ dissolves in water to make a solution which has a nominal concentration of 4.9×10^–3 M. Nominal meaning we pretend that the CaSO₄ remains as a molecular unit in water when we quote this number. It doesn't. What is it's equilibrium constant K_sp?

                           CaSO4(s)  Ca2+(aq) + SO42–(aq)

Initial conc.              ignore          0          0
x dissolves (solubility x = 4.9×10–3 M.)
at equilibrium concs are                   x          x

K_sp = [Ca²⁺][SO₄^2–] = x² = (4.9×10^–3)² = 2.4×10^–5.

In this next example, we will use a known K_sp value to determine the solubility of a salt.

The solubility product constant of PbCl₂ is 1.7×10^–5@ 25°C. What is the solubility of lead(II)chloride in water at 25°C?

                           PbCl2(s)  Pb2+(aq) + 2 Cl–(aq)
  I                          ignore          0          0
  C x dissolves (solubility x)              +x         +2x
  E                                          x          2x

Ksp = [Pb2+][Cl–]2 
    = x(2x)2 = 4x3 = 1.7×10–5.
 x3 = (1.7×10–5)/4
  x = (4.3×10–6)1/3
  x = 1.6×10–2 .

  Solubility = 1.6×10–2 M.

In this next example, we are going to use equilibrium ideas to determine if a precipitate will form.

Lets mix equal volumes of 2.0×10^–4 M AgNO₃ with 2.0×10^–4 M NaCl. Does a ppt form?

Since equal volumes of the two solutions are added together, the volume doubles and hence, each concentration is halved.

The only possible reaction between the four ions present here Ag⁺, Na⁺, Cl^–, NO₃^–, is:

AgCl(s) Ag⁺(aq) + Cl^–(aq)               (NaNO₃ is very soluble.)

Q = [Ag⁺][Cl^–] = (1.0×10^–4)(1.0×10^–4) = 1.0 × 10^–8.

K_sp = 1.7×10^–10. thus, Q > K therefore (rxn will proceed to left) ppt will form.

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Common Ion Effect

The solubility of an ionic compound in a solution which already contains one of the ions in that compound is reduced. This is the common ion effect.

Consider the following.

PbCl₂(s) Pb²⁺(aq) + 2 Cl^–(aq)

If we add some NaCl (or any other soluble chloride) we cause a stress on the equilibrium ([Cl^–] increases). LP states that the equilibrium will shift to the left to try to use up the extra chloride, i.e., more ppt will form.

Example

What is the solubility of PbCl₂ in 1.00 M HCl?

                     PbCl2(s)  Pb2+(aq) + 2 Cl–(aq)
I                                     0         1.00
C x dissolves (x=solubility)          +x        +2x
E                                     x         1.00 + 2x

K_sp = [Pb²⁺][Cl^–]² = 1.7×10 ^–5.

  x (1.00 + 2x)² = 1.7×10 ^–5.

Now let's use the small K approximation to simplify this and avoid using the quadratic equation to solve this. Since K is small, we assume x is small with respect to the initial amounts. Thus we assume that 1.00 + 2x can be approximated by just 1.00. Thus...

  x (1.00 M)² =  1.7×10 ^–5.

  x = 1.7×10 ^–5 M.

If no HCl or any other chloride had been present we would have determined the solubility to be 1.6×10^–2M. The solubility is significantly reduced due to common ion effect.

The common ion effect is a very useful way to deliberately reduce the solubility of slightly soluble ions. Consider Radon Ra⁺⁺(aq). This radioactive ion can be separated out of a mixture by adding an anion which will precipitate out with it such as the sulfate ion.

RaSO₄ (s) Ra²⁺(aq) + SO₄^2–(aq)    K_sp = [Ra²⁺][SO₄^2–] = 3.6 × 10^–11.

solubility of RaSO₄ in pure water is (3.6 × 10^–11)^1/2 or 6.0×10^–6M.

If we add enough H₂SO₄ to make a 1.00 M solution. we find

[Ra²⁺] = K_sp/[SO₄^2–] = (3.6 × 10^–11)/ 1.00 = 3.6 × 10^–11M.

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Effect of pH on Solubility

Many weakly soluble ionic compounds have solubilities which depend on the pH of the solution. A direct example is hydroxides since the OH^– ion is directly involved in the equilibrium constant. Other cases of pH dependence may not be quite so simple.

NOTE: pH is commonly defined as –log[H⁺]. This is not accurate. It should more accurately be defined as –log[a_H+], i.e., activity of H⁺ should be used. Click here for more info.

Example

Zinc hydroxide Zn(OH)₂ has K_sp=4.5×10^–17

In pure water:

Solubility is computed as follows:

                    Zn(OH)2(s)  Zn2+(aq) + 2OH–(aq)

I                                       0         0
C x dissolves                          +x       +2x
E                                       x        2x

K_sp = 4.5×10^–17 = x(2x)²

x = (4.5×10^–17 )/4]^1/3 = 2.2×10^–6 M.

the resulting pH is [OH^–] = 2x = 4.4×10^–6M therefore

pH = 14 – pOH
   = 14 – (–log( 4.4×10–6))
   = 8.64

This pH is the equilibrium pH resulting from dissolving the zinc hydroxide in pure water. LP states if we stress the system (by changing the pH) then the equilibrium will shift to reduce the stress.

if pH < 8.64 (more acidic) then [OH^–] decreases (rx shifts right to try to produce more). Solubility increases.

if pH > 8.64 (more basic) then [OH^–] increases (rx shifts left to try to use more). Solubility decreases.

Now what will happen in this example is we start again but this time, we try to dissolve the zinc hydroxide into a solution buffered at pH = 6.0.

[OH–] = antilog[ –(14 – pH)]
      = antilog[ –(14 – 6)] = 1.0×10–8.

Ksp   = [Zn2+][OH–]2 

[Zn2+] = Ksp/[OH–]2 = 4.5×10–17/(1.0×10–8)2 = 0.45 M. (= x).

Much larger solubility than than in pure water.

Here's a point! Tooth enamel is largely Ca₅(PO₄)₃OH (hydroxyapatite). What do you think will happen to your teeth if organic acids (say from the partial digestion of foods rich in sugar by the saliva and biota in your mouth) are allowed to stay in your mouth. (brush your teeth within 15 minutes of eating even one mouthful).

Any salt containing acidic or basic ions will have pH-dependent solubilities. Take for example,

 CaCO₃.  If the pH is sufficiently low (so that the HCO₃^–/CO₃^2– reaction has completely used up the carbonate.  In the distribution diagram for the carbonic acid system below, we see that at pH=8 there is (nearly) no carbonate (blue) left

(1)	CaCO3(s)		Ca2+(aq) + CO32–(aq)	Ksp(CaCO3)=3.36x10–9
(2)	CO32–(aq) + H3O+(aq)		HCO3–(aq) + H2O(l)	1/Ka2(H2CO3)=1/4.7x10–11                   = 2.2x1010
Sum	CaCO3(s) + H3O+(aq)		Ca2+(aq) + HCO3–(aq) + H2O(l)	Ksum

Ksum = Ksp(CaCO2)/Ka(HCO3–) = 3.36×10–9 / 4.7×10–11 = 71.5.  

The first thing to notice about this is that the second equilibrium is merely the K_a2(H₂CO₃) equation, written backwards.  At around pH=8 or lower, we can approximate that it proceeds to completion, using up almost all of the carbonate ion (See the distribution diagram above). Thus, we can assume that the concentration [CO₃^2–] is negligibly small.  Hence, adding the two equations together and cancelling out the CO₃^2– from the equation is a good approximation.  If the pH were lower than, say, 5 then we could say the same for the K_a1(H₂CO₃) equilibrium.

From the Summed equation, we see that, according to LeChatelier, if we add acid to the solution in equilibrium,  the overall equilibrium will shift right to use up some of the extra acid. This will cause the solubility of the CaCO₃ to increase.

Example:  Calculate the solubility of CaCO₃ in a buffered solution of pH=8.

Since the second equilibrium (above) goes to completion, we can use the summed equilibrium.

SUM:     CaCO3(s) + H3O+(aq) Ca2+(aq) + HCO3–(aq) + H2O(l)

I                   1×10–8          0          0 
C x dissolves:                     +x         +x
E                   1×10–8          x          x     

Remember that the solution is buffered so the H₃O⁺ concentration doesn't change (we assume).

Another example:  In this one, we can not assume that the intermediate concentrations are negligibly small.  We have to deal with both equilibria in full.  This complicates the mathematics as we will see below.

What is the solubility of CaF₂ in a solution buffered at:

        A: pH = 5.0? 
        B: pH = 3.0?
        C: pH = 1.0?

(1) CaF2 Ca2+ + 2F–	Ksp = 3.45×10–11
(2) HF H+ + F–	Ka = 6.3×10–4

A.  At pH=5, we can see from the diagram (or by comparing pKa to pH) that all the F^- from the dissolution of the CaF₂ remains in solution unchanged.  Therefore, this particular case is just a simple K_sp question.  No further calculations necessary.

               CaF2  Ca2+ + 2F–          
 I                         0      0
 C                         +x     +x
 E                         x      x

Ksp= 3.45×10-11 = x2   So x = 5.9×10-6

The solubility of CaF₂ at pH = 5 (or higher) is 5.9×10^-6 M.

B.  At pH=3, we can see from the distribution diagram that some of the F^– (BLUE) that came from the CaF₂ will be protonated into HF (RED). The ratio (read off the diagram or calculate using the Clausius Clapeyron equation) HF to F^– is approximately .61:.39 ~ 1.56 : 1, i.e.,

[HF] = 1.56[F^–].

Thus, we must deal with the two equilibria together with no assumptions possible. 

Now, according to the first equilibrium, the amount of HF produced is two times the amount of CaF₂ dissolved or twice the amount of Ca²⁺ produced.

[F]_total = 2[Ca²⁺]

The F^– that comes from the CaF₂ does not all stay in that form. Some of it becomes HF.   The total amount of fluorine however remains the same and can be represented by the relationship

[F]_total = [HF] + [F^–].

[HF] + [F^–] = 2[Ca²⁺]

(1.56 + 1)[F^–] = 2[Ca²⁺]

[F^–] = 0.781[Ca²⁺].

If s is the solubility, then [Ca²⁺] = s.

K_sp = [Ca²⁺][F^–]² = s(0.781s)² = 3.45×10^–11

s = 3.8×10^–4. Obviously, at this lower pH, the solubility of the CaF₂ is higher

---

C.  At pH=1, we can assume that all the F^– is now converted into HF and add equation 1 and 2 together.

(a)	CaF2 Ca2+ + 2F–	Ksp = 3.45×10–11
(b)	HF H+ + F–	Ka = 6.3×10–4
c = a–2b	CaF2 + 2H+ Ca2+ + 2HF	K = 3.45×10–11 x 1/(6.3×10–4)2    = 8.7×10–5

So now, our Ice table is made using this new final equation.

                    CaF2 + 2H+  Ca2+ + 2HF
       I                  0.1         0      0
       C                             +x    +2x
       E                  0.1         x     2x

and we now have:  K =  x(2x)2 =   8.7×10–5  (Buffered at pH = 1 means H+ conc doens't change)
                       0.12

                  x = 6.0×10–3

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We use these ideas of equilibrium and the application of LP to selectively alter the solubilities of ionic compounds in the qualitative analysis labs.

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Metal Sulfides

There are many Industrial applications were it is important to control the solubility of metal sulfides.  We can easily adjust the solubility of these ions using pH such that we can selectively precipitate Metal Sulfide solids from a mixture and effectively separate the metal ions as we filter off the precipitates in turn.

Metal II Sulfides: (Metal has oxidation state of +2; for example, M = Zn²⁺, Fe²⁺, ...)

1) MS(s)  M2+ (aq) + S2–(aq)

and

2) S2–(s) + H2O(l)  HS-(aq) +  OH–(aq)

However, the S^2– ion is a strong base (K_b~10⁵) and will react immediately to form HS^– and a hydroxide ion. So, the two the concentration of S^2– in solution is negligible and therefore, the actual dissolution process can be approximated as the sum of the reactions 1 and 2 above.

summ: MS(s) + H2O(l)  M2+ (aq) + HS–(aq) + OH-.

Thus, the true K_sp for the dissolution of a metal II sulphide is.

K_sp = [M²⁺][HS^–][OH^–]

Here we see that the addition of acid will use up OH^– and hence, shift the summed equilibrium to the right  thus dissolving more of the salt (MS).  Since the Solubility is higher in acid solution and quite low in base solution, it is often more convenient (and conventional) to rewrite the equation for the dissolution in an acidic solution.  We can use K_w = [H₃O⁺][OH^-] to do this.

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

We call such an equilibrium constant K_spa for solubility-product constant in acid.

we can determine the solubility s = [M²⁺] = [H₂S]

We can see that as the pH is lowered (higher H₃O⁺ concentration) the solubility of the metal sulfide increases.

For example:

If we add acid slowly, the FeS will dissolve first (pH = ca.3-4) since it's solubility is larger for a given [H⁺]. As we continue to add acid, the ZnS will eventually dissolve as well
(pH => –1).
If we had a solution containing Zn²⁺ and Fe²⁺ (both at 0.10M) we could selectively precipitate the Zn by buffering the pH to 2.38. At that pH, the solubility of FeS is exactly 0.1 while that of the ZnS is 7.0×10^-4. So, , assuming [H₂S] = 0.1M, at that pH, the ZnS would be mostly in the solid form but the FeS would be Just soluble.  If the pH were to raise above 2.38 or if the amount of H₂S were to be increased by even the tiniest amount, then some FeS would start to form ppt as well.

There are many methods of adding reagents to a mixture of ions to selectively separate out the individual components. Various types of reactions are covered in the Qualitative Analysis lab.

EXAMPLE:  You have a sample containing both Iron and Zinc ions, both at a concentration of 0.10 M.  The initial pH of the solution is 0, i.e., the [H⁺] = 1.0 M.  To what pH must you change the solution to get maximum separation of the iron and zinc ions if the H₂S nominal concentration is also 0.10 M?

First, the words "maximum separation" means we want to precipitate one of the ions as a salt while leaving the other ion in solution.  Since it is impossible to completely separate the ions, we look for the conditions that give us the best outcome possible.  This will occur when one of the ions is just barely in solution (at K_sp but with no solid yet formed), while the other is already mostly precipated.

We will be likely working in the low pH range since at high pH, the solubilities of both ions with sulfide is increased.

Hence, we'll use the K_spa.  setup. 

     MS(s) + 2H3O+(aq)  M2+ (aq) + H2S(aq) + 2H2O(l)
E             x              0.10       0.10

x is the H₃O⁺ concentration I need to make the equality true in the relationship

             Kspa =  [M2+][H2S]
                         x2

for FeS, we get

             6×102 =  [0.10][0.10]
                         x2

                  x = 4.1×10-3

for ZnS, we get

             3×10-2 =  [0.10][0.10]
                           x2

                  x = 0.57

To precipitate either or both of these ions, we need to lower the [H⁺] by slowly adding a base like NaOH.  Clearly, the Fe²⁺ will precipitate at a lower concentration of H⁺ than will the Zn²⁺.  So as we lower the concentration of the acid through 0.57, The Zn²⁺ will begin to precipitate.  when we reach a value of [H⁺] = 4.1×10^-3 most of the Zn²⁺ will have already precipitated and none of the Fe²⁺.  At this point, one more drop of base would make the FeS start to precipitate so we don't go there.  We have reached the point of maximum separation.

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Complex Ion Equilibria

Definition:

Complex Ion:

An ion consisting of a metal ion surrounded by ligands.

Ligand:

A molecule or ion having alone-pair of electrons which can be 'donated' to a metal ion to form a covalent bond.

Some common ligands include H₂O, NH₃, Cl^–, CN^–.

Coordination Number

The number of ligands attached to a central cation.

Metal ions in Solution form complexes by covalently bonding to some number of ligands. The bond is a special kind of bond where the ligand donates one or more of it's lone-pairs of electrons to one of the empty orbitals in the d-shell of the metal ion. These interactions can almost always be written as an equilibrium with all the requisite properties of equilibrium being valid. The reactions are always written so that one mole of the complex is the only product. Thus, these particular equilibrium constants are called formation constants or stability constants. for example, if we mix silver ions with ammonia we can observe the following two complexation reactions.

1)            Ag+(aq)   +   NH3(aq)     Ag(NH3)+(aq)	Kcx1 = 2.1×103.
2) Ag(NH3)+(aq)   +   NH3(aq)      Ag(NH3)2+(aq)	Kcx2 = 8.2×103.

Where K_cx1 and K_cx2 are the formation constants for the two complexes Ag(NH₃)⁺(aq) and Ag(NH₃)₂⁺(aq), respectively.

If a large enough excess of NH₃ is present we can consider that the two reactions essentially go to completion and hence, the intermediate concentrations of Ag(NH₃)⁺(aq) are very small (We check this later in the calculation). The overall reaction would then be

Ag+(aq)   +   2 NH3(aq)        Ag(NH3)2+(aq)

Kf = Kcx12 =  Kcx1×Kcx2 = 1.7×107

we have               Ag(NH3)+(aq) +   NH3(aq)  Ag(NH3)2+(aq)

I                         0            1.0            5.0×10–4 
C x reacts (backwards)   +x                             -x
E                         x            1.0            5.0×10–4 – x

assume x is small cf. 5.0×10–4

                     Ag+(aq) + NH3(aq)  Ag(NH3)+(aq) 

I                   0          1.0           6.1×10–8
C                  +x          +x            -x
E                   x          1.0+x           6.1×10–8 -x
                              (assume x is very small again)

               AgCl(s)  Ag+(aq) + Cl–(aq)     Ksp = 1.6×10–10 

     Ag+(aq) + NH3(aq)  Ag(NH3)+(aq)          Kcx1 = 2.1×103

Ag+(NH3)(aq) + NH3(aq)  Ag(NH3)2+(aq)         Kcx2 = 8.2×103

AgCl(s) + 2 NH3(aq)  Ag(NH3)2+(aq) + Cl–(aq)

K for this reaction is the product of the three equilibrium constants

                 AgCl(s) + 2 NH3(aq)  Ag(NH3)2+(aq) + Cl–(aq)

I                             10.0 M            0             0
C x dissolves                 -2x               +x            +x
E                             10.0 – 2x         x             x

take square root of both sides:

             AgCl(s)  Ag+(aq) + Cl–(aq)     Ksp = 1.6×10–10 

I                            0         0
C                           +x        +x  
Equilibrium                  x         x