Solubility Equilibria

In these notes, we look at chemical equilibrium as it applies to salts that have a very low solubility in water.  As you may remember from General Chemistry I, when salts dissolve in water, they split into their component ions -- a process we call dissociation.  For example, when table salt dissolves in water, we can write the reaction as

NaCl(s)   ---------->   Na+(aq)   +   Cl-(aq)

Because NaCl is freely soluble in water, we don't usually saturate the solution so an equilibrium between the solid salt and its ions is not established.  Only if you put so much salt in water that it would not all dissolve would you have the following equilibrium occurring:

NaCl(s)   <---------->   Na+(aq)   +   Cl-(aq)

The only difference in these two equations is the bidirectional arrow in the second equation as compared to the unidirectional arrow in the first.  The second equation indicates the condition in which Na+ ions and Cl- recombine to form solid NaCl at the same rate that NaCl dissolves to form Na+ ions and Cl-.

Normally, when we consider the equilibria of solid salts with their ions in aqueous solution, it is with salts of a very low solubility.  As an example, consider barium chromate

BaCrO4(s)   <---------->   Ba2+(aq)   +   CrO42-(aq)

Recall that an equilibrium constant is written as a fractional expression with the substances on the right on top and the substances on the left on the bottom, with each substance raised to a power equal to its coefficient.  With this in mind, we would write the concentration-based equilibrium constant as

                 [Ba2+]eq [CrO42-]eq
KC   =     ------------------------
                     [BaCrO4]eq 

However, we have seen that pure solids and pure liquids are omitted from the equilibrium constant expression.    One justification for this is that solids and liquid have an essentially constant concentration.  If we multiply both sides of the KC expression by [BaCrO4] we get

[BaCrO4] KC   =   [Ba2+]eq [CrO42-]eq 

But [BaCrO4] and KC are both constants, and a constant multiplied by a constant is just another constant.  If we let KSP be the new constant, that is,

KSP  =  [BaCrO4] KC

then we have

KSP  =  [Ba2+]eq [CrO42-]eq

KSP is referred to as the "solubility product" or sometimes "ion product constant" because it is obtained by multiplying the concentrations of the two ions.  Of course, if one of the ions has a coefficient other than 1, then that ion must be raised to a power.  For example, consider the salt barium fluoride:

BaF2(s)   <---------->   Ba2+(aq)   +   2F-(aq)

The solubility product expression for this salt is

KSP  =  [Ba2+]eq [F-]eq2 

Because the BaF2 is a solid, it does not appear in the equilibrium constant expression.  The justification for this has been discussed previously in these notes (when BaCrO4 was considered) and also in the notes on acid / base equilibria (to explain omitting H2O).  There will be no need to do this analysis each time.  From now on, we just know to multiply the concentrations of the two ions, raising each to the power of its coefficient in the equilibrium reaction.  We will never divide by the concentration of solid salt.

From the examples with BaCrO4 and BaF2 we see that the KSP expression for a slightly soluble salt can be written by inspection, referring to its formula.  In order to do calculations involving solubility equilibria, we will need to have numerical values for the KSP's of the salts we are working with.  Fortunately, such values are routinely published, and most general chemistry textbooks will include a table of KSP values.  Like any equilibrium constant, KSP's are temperature dependent.  The values published in general chemistry textbooks are usually for 25 oC.  In this course, we will always do our solubility equilibria calculations at 25 oC, so we can use the table of KSP values in working our problems.

Example Problem:  What is the molar solubility of BaCrO4?

Discussion and Answers

We can set up an "ICE" table (see previous notes on chemical equilibrium in general, and the notes on acid / base equilibria for detailed discussion of "ICE" tables)

The equilibrium reaction is

BaCrO4(aq) <----------> Ba2+(aq) + CrO42-(aq)

For each mole of BaCrO4 that dissolves, one mole of Ba2+ and one mole of CrO42- is produced.  Therefore, either of these two ions could be used to represent the molar solubility of BaCrO4.  That is,

Molar Solubility = [Ba2+]eq =  [CrO42-]eq  

So we need to find the equilibrium concentrations of the ions in a saturated BaCrO4 solution.  That's where the "ICE" table comes in.  We can construct the following table relating the Initial concentrations, the Changes in concentrations and the Equilibrium concentrations.

 

[BaCrO4]

Ba2+

CrO42-

I

-----

0 0

C

-----

+X +X

E

-----

X X

In the above table, nothing is filled in for BaCrO4 since it is omitted from the equilibrium constant expression.  An initial concentration of zero is entered for Ba2+ and CrO42- because these ions should not be present in distilled water.  Before the BaCrO4 dissolves, these ions are not present.  We know that the BaCrO4 will dissolve to a small extent, and in so doing, will generate Ba2+ and CrO42- ions.  From the salt's formula, we can see that these ions must be produced in equal molar amounts.  We let X represent the increase in concentration of Ba2+ and CrO42- caused by the dissolving of the salt.  This gives us an increase of +X for both Ba2+ and CrO42- and an equilibrium concentration of X for each, since the initial concentration was 0.  The equilibrium concentrations are obtained by adding the change in concentration to the initial concentration.  Here, we have 0 + X = X.

Now, we make use of the solubility product expression:

KSP = [Ba2+]eq [CrO42-]eq = (X)(X) = X2 = 1.2 x 10-10

In the above equation, the value 1.2 x 10-10 is the KSP for BaCrO4 at 25 oC.  Taking the final equality in the above chain,

X2 = 1.2 x 10-10

we can take the square root of both sides to get

X = 1.1 x 10-5

Therefore,

[Ba2+]eq = [CrO42-]eq = molar solubility = X = 1.1 x 10-5 mol L-1

Example Problem: Calculate the solubility of BaCrO4 in g/L and mg/L.

Discussion and Answers

This is a logical follow-up to the previous problem.  We now know the number of moles of BaCrO4 per liter of solution, and it is a simple matter to convert moles to grams, and to convert grams to milligrams.

            mol             253.3 g                     g
1.1 x 10-5 ---------   x    ---------   =   2.8 x 10-3 ----
             L                mol                       L

            g           1 mg                mg
2.8 x 10-3 ----   x   -----------   =   2.8 ----
            L         1 x 10-3 g             L

Notice how low the solubility is.  A liter of solution can hold only 2.8 mg of BaCrO4.  Solubility equilibria calculations are usually carried out for salts of very low solubility.  These are referred to as "slightly soluble salts" or sometimes, even "insoluble" salts.  Of course, even "insoluble" salts dissolve in water to a very small extent.

Example Problem: Calculate the molar solubility of BaCrO4 in a 0.10 M Ba(NO3)2 solution.

Discussion and Answers

This problem involves what is known as the common ion effect.  A salt which is only slightly soluble in water is even less soluble in an aqueous solution containing an ion that is common to the salt.  In this example, Ba2+ is the common ion.  It is found in both the salt (BaCrO4) and the solution in which we are trying to dissolve the salt.  The Ba(NO3)2 is a freely soluble salt and dissolves without establishing an equilibrium:

Ba(NO3)2(s)   ==========>   Ba2+(aq)   +   2NO3-(aq)

If 0.10 mole of Ba(NO3)2 is added to enough water to make 1 L of solution, there will be 0.10 mole of Ba2+ ions and 0.20 mole of NO3- ions in the solution.  This is clear because the formula of the salt (and the chemical equation shown above for the dissolving of Ba(NO3)2) shows that 1 mole of Ba(NO3)2 produces 1 mole of Ba2+ ions and 2 moles of NO3- ions.  So for any quantity of Ba(NO3)2 that is dissolved in water, the number of moles of Ba2+ ions will be equal to the number of moles of solid Ba(NO3)2 that was dissolved, and the number of moles of NO3- ions will be twice the number of moles of Ba2+ ions. The ICE table for the dissolving of Ba(NO3)2 in water is as follows:

 

[Ba(NO3)2]

Ba2+

NO3-

I

-----

0 0

C

-----

+0.10 +0.20

E

-----

0.10 0.20

Here, the ICE table does not have an unknown variable X in it, because the extent of ionization of the Ba(NO3)2 is known to be 100%.  So if we know how many moles of Ba(NO3)2 are used for every liter of solution, then we know the concentration of Ba2+ and NO3- ions.  Now consider dissolving BaCrO4 in this already prepared Ba(NO3)2 solution.  Barium chromate, BaCrO4 is a slightly soluble salt, and establishes its usual equilibrium in an aqueous environment:

BaCrO4(s)   <---------->   Ba2+(aq)   +   CrO42-(aq)

But this time, there are Ba2+ ions in the solution even before the BaCrO4 dissolves.  So when the BaCrO4 dissolves, the Ba2+ ions produced will be in addition to those that were already there from the dissociation of Ba(NO3)2.  One way to explain the reduced solubility of a slightly soluble salt in the presence of a common ion is through Le Chatelier's principle.  Because there is additional Ba2+ present (from the dissociation of Ba(NO3)2), the equilibrium will be shifted to the left, which is the side with solid BaCrO4.  Therefore, a greater amount of BaCrO4 will exist in the solid form than if the excess Ba2+ had not been there.  By doing an ICE table calculation for this system, we can calculate the solubility in the presence of the common ion.  We usually find quite dramatic reductions in solubility brought about the presence of a common ion.

 

[BaCrO4]

Ba2+

CrO42-

I

-----

0.10 0

C

-----

+X +X

E

-----

0.10 + X X

 In the above ICE table, 0.10 has been entered for the initial concentration of Ba2+.  This is the molar concentration of Ba2+ that is already present in the solution from the dissociation of Ba(NO3)2.  The BaCrO4 then dissociates to a small extent in this solution, increasing the Ba2+ and CrO2- concentrations by equal amounts -- represented by X in the ICE table.  At equilibrium, the concentration of Ba2+ is 0.10 + X.  That is, the equilibrium concentration of Ba2+ is equal to the sum of its initial concentration and the amount by which the BaCrO4 has increased this concentration.  The initial concentration of CrO42- is zero.  The Ba(NO3)2 does not provide any CrO42-.  The BaCrO4 is the sole source of CrO42-, so before the solid BaCrO4 dissolves, there are no CrO42- ions in solution.  The dissociation of BaCrO4 increases the concentration of CrO42- by X, and since the initial concentration was 0, the equilibrium concentration is 0 + X, which is X.  We can now substitute the equilibrium concentrations into the equilibrium constant expression:

KSP = [Ba2+]eq [CrO42-]eq = (0.10 + X)(X) = 1.2 x 10-10

An exact solution of this equation requires the quadratic formula.  We start by multiplying (0.10 + X) by X to get

X2 + 0.10X = 1.2 x 10-10

Then, by subtracting 1.2 x 10-10 from both sides, we get

X2 + 0.10X - 1.2 x 10-10 = 0

This has the form of a standard quadratic equation,

aX2 + bX + c = 0

for which the solutions are

     -b (b2 - 4ac)
X = -----------------
          2a

In solving this equation, we can ignore the negative root, since this would lead to a negative equilibrium concentration -- an impossible situation.  Concentrations can never be negative.  Comparing the generic form of the standard quadratic equation to our specific equation, we see that

a =1

b = 0.10

c = -1.2 x 10-10

and the solution to our quadratic equation is

     -0.10 + ( 0.102 - 4(1)(-1.2 x 10-10) )
X = ----------------------------------------
                   2(1)

Inside the square root, we can square the 0.10 to get 0.010 and we can multiply the -4, the 1 and the -1.2 x 10-10 to get +4.8 x 10-10.  We now have

     -0.10 + (0.010 + 4.8 x 10-10)
X = -----------------------------------
                2(1)

Carrying out the addition inside the square root, and the multiplication in the denominator, we have

     -0.10 + (0.01000000048)
X = -------------------------------
                 2

Taking the square root gives

     -0.10 + 0.1000000024
X = ------------------------
             2

And finally, performing the addition in the numerator and then dividing by 2, we have

     0.0000000024
X = ---------------- = 0.0000000012 = 1.2 x 10-9
          2

We now have the molar solubility of BaCrO4 in the 0.10 M Ba(NO3)2 solution:

molar solubility = X = [CrO42-]eq = 1.2 x 10-9 mol/L

Notice that only the CrO42- concentration can be used to represent the molar solubility of BaCrO4.  Since Ba2+ ions are present from another source -- the Ba(NO3)2 -- the concentration of Ba2+ ions does not indicate how much BaCrO4 dissolved in the solution.  From the ICE table, we see that the equilibrium concentration of Ba2+ ions is as follows:

[Ba2+]eq = 0.10 + X = 0.10 + 1.2 x 10-9 = 0.1000000012 mol/L

For all practical purposes, [Ba2+]eq = 0.10 mol/L.  The BaCrO4 has a very low solubility even in pure water, and this solubility is even lower in the presence of the common ion.  The amount of Ba2+ that is contributed by the dissociation of BaCrO4 is negligible compared to that which is already there from the dissociation of Ba(NO3)2.  It's like adding a drop of water to a swimming pool.

If solving the quadratic equation seemed like a lot of work, you will be glad to know that there is a shortcut we can take to avoid it.  Now that you've seen it worked the long way around, let's go back to the original equation from the ICE table and see how much easier it can be.

 KSP = [Ba2+]eq [CrO42-]eq = (0.10 + X)(X) = 1.2 x 10-10

Even if you had not previously seen the quadratic solution to this equation, you can still determine that X is going to very small in comparison to 0.10.  We previously worked out the molar solubility of BaCrO4 in pure water and found it to be 1.1 x 10-5 mol/L.  This is already small in comparison to 0.10 mol/L, and we expect it to decrease even further in the presence of the common ion.  If X is extremely small in comparison to 0.10, then

0.10 + X ≈ 0.10

With this in mind, the equation

(0.10 + X)(X) = 1.2 x 10-10

reduces to

0.10X = 1.2 x 10-10

Now, we only have to divide both sides by 0.10 to arrive at our answer!

     1.2 x 10-10
X = ------------ = 1.2 x 10-9
       0.10

This is the same answer we arrived at using the quadratic formula, but this time we got the answer with only a fraction of the work!

Before closing this discussion, let's note how much the common ion suppressed the solubility of BaCrO4:

Solubility in pure water:  1.1 x 10-5 mol/L

Solubility in 0.10 M Ba(NO3)2: 1.2 x 10-9 mol/L

Compare these two solubilities:

1.1 x 10-5
---------- = 9.2 x 103
1.2 x 10-9

The BaCrO4 is more than 9000 times more soluble in pure water than in a 0.10 M Ba(NO3)2 solution.  Or conversely, the BaCrO4 is more than 9000 times less soluble in 0.10 M Ba(NO3)2 than in pure water. 

Example Problem:  Calculate the solubility of BaCrO4 in 0.10 M Ba(NO3)2 in g/L and μg/L.

Discussion and Answers

Make use of the molar solubility in the previous problem.  Convert moles per liter to grams per liter and then convert grams per liter to micrograms per liter.

           mol       253.3 g                 g
1.2 x 10-9 ----  x  ---------  =  3.0 x 10-7 ---
            L         1 mol                  L

            g         1 μg              μ
3.0 x 10-7 ----  x  ---------  =  0.30 ---
            L       1 x 10-6 g          L

Less than one microgram of BaCrO4 dissolves in a liter of 0.10 M Ba(NO3)2 solution.  In comparison, 2.8 milligrams would dissolve in a liter of pure water.

The previous two examples have shown that the solubility of a slightly soluble salt (BaCrO4) is drastically reduced in the presence of a common ion (Ba2+).  Earlier, it was mentioned briefly that this could be explained on the basis of Le Chatelier's  Principle.  This principle states that if a chemical equilibrium is subjected to a stress, it will adjust in such a way as to counteract or minimize that stress.  The equilibrium between solid BaCrO4 and its ions in an aqueous environment is represented by the the equation

BaCrO4(s)   <----------->   Ba2+(aq)   +   CrO42-(aq)

In a 0.10 M Ba(NO3)2 we can think of the "stress" as an overabundance of Ba2+ provided by the Ba(NO3)2.  In response to this, the BaCrO4 dissolves to a smaller extent, so that not as much Ba2+ is brought into solution as it would be in distilled water, where no Ba2+ is initially present.  But of course, chemical equilibria don't have feelings, and don't experience stress.  That's just a convenient way for us to remember it.  So what is the physical explanation for this phenomenon we attribute to Le Chatelier's Principle?  As the BaCrO4 dissolves in the solution, it contributes CrO42- ions to the solution.  The rate of the reverse reaction, in which Ba2+ and CrO42- combine to form solid BaCrO4, depends on the collision frequency between Ba2+ ions and CrO42- ions.  In any given time interval, a CrO42- ion has a greater chance of colliding with a Ba2+ ion if there is a greater concentration of Ba2+ ions.  So having an alternate source of Ba2+ ions causes the reverse reaction to proceed at a faster rate, meaning equilibrium between solid BaCrO4 and its ions occurs at a lower CrO42- concentration.  That is, the BaCrO4 is less soluble in the presence of excess Ba2+.  In pure water, BaCrO4 is the sole source of both Ba2+ ions and CrO42- ions.  Without the additional Ba2+, the CrO42- concentration must be higher in order for the rate of the reverse reaction to be fast enough to compete with the forward reaction.  So equilibrium is established with a higher concentration of CrO42- in pure water than in the 0.10 M Ba(NO3)2 solution.  Therefore, more BaCrO4 dissolves in pure water than in 0.10 M Ba(NO3)2.

We have been considering how Le Chatelier's Principle reduces the solubility of a slightly soluble salt in the presence of a common ion.  It is also possible to enhance the solubility of a slightly soluble salt.  This can be done if there is a complexing agent that binds to one of the ions in solution and therefore, reduces the concentration of the free ion.  From the perspective of Le Chatelier's Principle, the "stress" is the removal of one of the ions from the solution, and to counteract that stress, the salt dissolves to a greater extent to replace the ions that have been removed by the complexing agent.  As an example, we consider AgCl.  In the following 3 problems, we look at the solubility of AgCl in pure water, a 0.10 M NaCl solution and a 0.10 M NH3 solution.  The NaCl will provide Cl- ions, which will reduce the solubility of AgCl because the Cl- is a common ion.  The NH3 will serve as a complexing agent for Ag+, reducing the concentration of free Ag+ ions in the solution, thus causing additional AgCl to dissolve (to replace the Ag+ ions that have been removed).

Example Problem: Calculate the molar solubility of AgCl in water.  Note that KSP for AgCl is 1.8 x 10-10.

Discussion and Answers

The chemical equation for the equilibrium of solid AgCl with its ions in solution is

AgCl(s)   <---------->   Ag+(aq)   +   Cl-(aq)

The ICE table for this reaction is as follows:

 

[AgCl]

[Ag+]

[Cl-]

I

-----

0 0

C

-----

+X +X

E

-----

X X

We don't need to make any entries for AgCl, since it is a solid, and will be omitted from the equilibrium constant expression.  The AgCl is being dissolved in distilled water, so the initial concentration is zero for both Ag+ and Cl-.  The AgCl dissolves to a small extent, increasing the Ag+ and Cl- concentrations by equal amounts, represented by X in the ICE table.  Since the initial concentrations were 0, the equilibrium concentrations are X.  The equilibrium concentrations can now be substituted into the KSP expression for this salt:

KSP = [Ag+]eq [Cl-]eq = (X)(X) = X2 = 1.8 x 10-10

Using the last equality in the chain,

X2 = 1.8 x 10-10

take the square root of both sides to get

X = 1.3 x 10-5 mol/L

Each mole of AgCl produces one mole of Ag+ and one mole of Cl- when it dissolves.  Therefore, both of these ions are representive of the molar solubility of AgCl.

Molar solubility = [Ag+]eq = [Cl-]eq = X = 1.3 x 10-5 mol/L

Example Problem: Calculate the molar solubility of AgCl in a 0.10 M NaCl solution.  Note that KSP for AgCl is 1.8 x 10-10.

Discussion and Answers

The NaCl is a freely soluble salt.  A concentration of 0.10 M is far from saturated for NaCl, so no equilibrium will exist between solid NaCl and its ions.  Rather, the entire solid mass of NaCl will completely dissociate in the solution, and only the ions will be present.  If the amount of solid NaCl added to the water is enough to provide 0.10 moles of NaCl for each liter of solution, then the concentration of Na+ and Cl- in the solution will both be 0.10 M.  Shown below is the ICE table for the dissociation:

 

[NaCl]

[Na+]

[Cl-]

I

-----

0 0

C

-----

+0.10 +0.10

E

-----

0.10 0.10

In the above ICE table, nothing is entered for NaCl, since it is a solid, and would not appear in the equilibrium constant expression.  The NaCl is being dissolved in distilled water, so the initial concentrations of Na+ and Cl- are both zero.  The salt fully dissolves, providing 0.10 moles each of Na+ and Cl- per liter of solution.  At "equilibrium" (though that's not really what we have here) the concentrations of Na+ and Cl- are both 0.10 M.  Since we will now be dissolving AgCl in this 0.10 M NaCl solution, the Cl- is a common ion.  Its presence will suppress the solubility of AgCl, so it will be less soluble in this solution than it was in pure water.  When the AgCl is dissolved, it establishes the usual equilibrium between solid AgCl and its ions in solution:

AgCl(s)   <---------->   Ag+(aq)   +   Cl-(aq)

The AgCl is the sole source of Ag+ ions, but Cl- will be present even before the AgCl dissolves due to the NaCl in the solution.  In this system, only the Ag+ concentration is representative of the molar solubility of AgCl.  The concentration of Cl- will not be indicative of the amount of AgCl that dissolved, since the majority of the Cl- ions will have come from NaCl, not AgCl.  Shown below is the ICE table for the dissolving of AgCl in a 0.10 M NaCl solution:

 

[AgCl]

[Ag+]

[Cl-]

I

-----

0 0.10

C

-----

+X +X

E

-----

X 0.10 + X

In the above ICE table, nothing is entered for AgCl, since it is a solid and is omitted from the equilibrium constant expression.  Since the AgCl is the sole source of Ag+, the initial concentration of Ag+ is 0.  For the Cl-, the initial concentration is 0.10 M, as provided by the NaCl.  When the AgCl dissolves, the Cl- that it generates will be in addition to the Cl- that is already there from the dissociation of NaCl.  Substituting the equilibrium concentrations into the KSP expression for AgCl, we have

KSP = [Ag+]eq [Cl-]eq = (X)(0.10 + X) = 1.8 x 10-10

Recall that the molar solubility of AgCl in distilled water was only 1.3 x 10-5 mol/L.  Here, the molar solubility will be even lower, because of the common ion.  The variable X will be very small in comparison to 0.10.  That is,

0.10 + X ≈ 0.10

With this in mind, the equation

(x)(0.10 + X) = 1.8 x 10-10

reduces to

0.10X = 1.8 x 10-10

and dividing both sides by 0.10, we have

     1.8 x 10-10
X = ----------- = 1.8 x 10-9
       0.10

Therefore,

Molar solubility = [Ag+]eq = X = 1.8 x 10-9 mol/L

Comparing this to the molar solubility in pure water (1.3 x 10-5 mol/L), we see that the solubility has been reduced by a factor of more than 7000.

1.3 x 10-5 mol/L
---------------- = 7.2 x 103
1.8 x 10-9 mol/L

Example Problem:  Calculate the molar solubility of AgCl in 0.10 M NH3.  Note that the KSP for AgCl is 1.8 x 10-10 and the formation constant for Ag(NH3)2 is 1.6 x 107.

Discussion and Answers

There are two equilibria we must consider in this system.  First, when the AgCl dissolves, it establishes its usual equilibrium:

AgCl(s)   <---------->   Ag+(aq)   +   Cl-(aq)

and second, the Ag+ ions react with the NH3 to form the complex ion [Ag(NH3)2]+:

Ag+(aq)   +   2NH3(aq)   <---------->   [Ag(NH3)2]+(aq)

The net effect of these two reactions is obtained by adding them together.  When reactions are added, their equilibrium constants are multiplied.  Here is what the chemical arithmetic looks like:

AgCl(s)   <---------->   Ag+(aq)   +   Cl-(aq)     KSP = 1.8 x 10-10

Ag+(aq)   +   2NH3(aq)   <---------->   [Ag(NH3)2]+(aq)     Kf = 1.6 x 107
------------------------------------------------------------------------
AgCl(s)  +  2NH3(aq)  <-------->  [Ag(NH3)2]+(aq)  +  Cl-(aq)  K = 2.9 x 10-3

In the above addition, the equilibrium constant for the second reaction is denoted Kf to indicate that it is a formation constant -- the term for the equilibrium constant of a reaction in which a complex ion is formed.  In adding the reactions, the Ag+ cancels out because it is both a product (first reaction) and a reactant (second reaction).  Now that we have the overall reaction and its equilibrium constant, we can construct an ICE table and substitute the equilibrium concentrations into the equilibrium constant expression.

  [AgCl] [NH3] [Ag(NH3)2]+ [Cl-]

I

--- 0.10 0 0

C

--- -2X +X +X

E

--- 0.10 - 2X X X

In the above ICE table, nothing has been entered for AgCl, since it is a solid and is omitted from the equilibrium constant expression.  Our starting concentration of NH3 is 0.10 M, and there is no [Ag(NH3)2]+ or Cl- in solution before the AgCl dissolves.  As the AgCl reacts with the NH3, the concentration of NH3 decreases and the concentrations of [Ag(NH3)2]+ and Cl- increase.  Therefore, the change in NH3 concentration is entered as a negative quantity, and the changes in [Ag(NH3)2]+ and Cl- concentrations are entered as positive quantities.  Notice that the change in NH3 concentration is entered as -2X rather than just -X.  This is in keeping with the stoichiometry of the balanced equation.  The equation shows that 2 moles of NH3 react to form 1 mole each of [Ag(NH3)2]+ and Cl-.  When filling out an ICE table, it is important that the concentration changes shown in the ICE table reflect the stoichiometry of the balanced chemical equation.

Since the balanced chemical equation is

AgCl(s)  +  2NH3(aq)  <-------->  [Ag(NH3)2]+(aq)  +  Cl-(aq)

the equilibrium constant expression will be

     [ [Ag(NH3)2]+ ]eq [Cl-]eq 
K = ------------------------- = 2.9 x 10-3
              [NH3]2eq 

We now substitute the equilibrium concentrations from the ICE table into this equation:

       (X)(X)
K = ------------- = 2.9 x 10-3
    (0.10 - 2X)2

Notice that the expression involving X is a perfect square.  There is X2 on top and (0.10 - 2X)2 on the bottom.  We can avoid having to solve a quadratic equation by simply taking the square root of this expression and the numerical value of K.

     X
----------- = √ (2.9 x 10-3)
(0.10 - 2X)

Note that √ (2.9 x 10-3) has two values: +5.4 x 10-2 and -5.4 x 10-2.  Therefore, the quantity will be left in the form √ (2.9 x 10-3) until it is time to calculate the final answer.  At that time, we will need to consider a "Case 1" in which the positive square root is used and a "Case 2" in which the negative square root is used.  Only one of these two cases will yield a value of X that falls within acceptable limits. More on that later.

Multiplying both sides of the equation we now have by (0.10 - 2x) gives

X = (0.10 - 2X) √ (2.9 x 10-3)

Carrying out the multiplication, we have

X = (0.10) √ (2.9 x 10-3) - (2X) √ (2.9 x 10-3)

and transposing the X-containing term from the right side to the left side gives

X + (2X) √ (2.9 x 10-3) = (0.10) √ (2.9 x 10-3)

On the left side, we can factor out the X to get

X ( 1 + 2√ (2.9 x 10-3) ) = (0.10) √ (2.9 x 10-3)

The final step in isolating X is to divide by ( 1 + 2√ (2.9 x 10-3) ):

     (0.10) √ (2.9 x 10-3)
X = -------------------------
     ( 1 + 2√ (2.9 x 10-3) )

Now the time has come to replace √ (2.9 x 10-3) with its two possible values.

Case 1: √ (2.9 x 10-3) = 5.4 x 10-2

     (0.10)(5.4 x 10-2)
X = ----------------------
     ( 1 + 2(5.4 x 10-2) )

Carrying out the multiplications in both the numerator and the denominator, we have

       5.4 x 10-3
X = -----------------
     1 + 1.08 x 10-1

Finally, we carry out the addition in the denominator and divide into the numerator to obtain the final answer for Case 1:

      5.4 x 10-3
X = ------------ = 4.9 x 10-3
       1.108

Case 2: √ (2.9 x 10-3) = -5.4 x 10-2

     (0.10)(-5.4 x 10-2)
X = -----------------------
     ( 1 + 2(-5.4 x 10-2) )

Carrying out the multiplications in both the numerator and the denominator, we have

         -5.4 x 10-3
X = --------------------
      1 + (-1.08 x 10-1)

Finally, we carry out the addition in the denominator and divide into the numerator to obtain the final answer for Case 2:

      -5.4 x 10-3
X = -------------- = -6.1 x 10-2
        0.892

Now we have two different solutions for X, and we must decide which one is correct.  We have to look at the ICE table and decide on the range of acceptable values for X.  From the ICE table we note the following:

[NH3]eq = 0.10 - 2X

[ [Ag(NH3)2]+ ]eq = [Cl-]eq = X

Since none of the equilibrium concentrations can be negative, X can not be negative.  The equilibrium concentration of both [Ag(NH3)2]+ and Cl- are equal to X, so this places a lower limit of 0 on X.  Now look at NH3.  This places an upper limit on X, because a value of X larger than 0.05 will result in a negative concentration for NH3.  Therefore, the range of acceptable values for X is

0 < X < 0.05

Anything outside of this range must be rejected as physically impossible.  Our two calculated values of X are 4.9 x 10-3 (Case 1) and -6.1 x 10-2 (Case 2).  Case 1 gives an acceptable value for X.  The value 4.9 x 10-3 is 0.0049 in floating point, and

0 < 0.0049 < 0.05

as required.  In this particular problem, it is pretty clear which value of X to choose because one of the values of X is negative.  The variable X is almost always equal to one of the equilibrium concentrations in the ICE table, so when you see a negative value of X, it is a pretty safe bet that you should reject it.  Notice also that in this particular problem, we ended up rejecting the negative square root.  It may be tempting to conclude that we always reject the negative square root and go with the positive square root.  But be aware that what we are rejecting is the negative value of X, not the negative square root.  In some problems, the "negative root" may yield a positive (and acceptable) value of X.  In general, we must check both cases to see which one generates an acceptable answer.

In solving our ICE table equation -- which, in its original form, was

       (X)(X)
K = ------------- = 2.9 x 10-3
    (0.10 - 2X)2

we had the good fortune of finding that the expression for X was a perfect square, and we could simply take the square root of the equation to avoid solving a quadratic equation.  But perhaps you were wondering if we could have used the same approximation technique we used earlier.  That is, assuming that

2X << 0.10

(<< means "much much less than") so that

0.10 - 2X ≈ 0.10

With that approximation, the equation would reduce to

       (X)(X)
K ≈ ------------- ≈ 2.9 x 10-3
      (0.10)2

so that

X2 ≈ (0.10)2 (2.9 x 10-3) = (0.010)(2.9 x 10-3) = 2.9 x 10-5

and

X = 5.4 x 10-3

This differs somewhat from the correct value of 4.9 x 10-3.  So why does the approximation fail in this case?  Recall the reaction and its equilibrium constant:

AgCl(s)  +  2NH3(aq)  <-------->  [Ag(NH3)2]+(aq)  +  Cl-(aq)  K = 2.9 x 10-3

The equilibrium constant is a little too large for the approximation to work well.  The larger the value of K, the more the equilibrium favors the right side of the reaction.  If there is significant reaction, then a significant amount of the reactants will be consumed.  The amount of NH3 consumed in the reaction is not negligible in comparison to the amount originally present.  Therefore, in the quantity

0.10 - 2X

the 2X is not negligible in comparison to 0.10.  Omitting the 2X changes the equation enough that the final answer deviates from the correct answer.  If we had not been fortunate enough to have a perfect square, we would have been forced to use the full quadratic formula to get the correct answer.  While approximation techniques can save a lot of time (and work!) you must be sure they are valid if you use them.  In all cases, it is a good idea to check your answer for consistency.  Let's do that now for the current problem:

We found that X = 0.0049.  Based on the ICE table, we have the following equilibrium concentrations:

[ [Ag(NH3)2]+ ]eq = [Cl-]eq = X = 0.0049 mol/L

[NH3]eq = 0.10 - 2(0.0049) = 0.10 - 0.0098 = 0.0902 mol/L

Now try substituting these concentrations back into the equilibrium constant expression and see if you get the original K value of 2.9 x 10-3.

{ [Ag(NH3)2]+ ]eq [Cl-]eq     (0.0049)(0.0049)     0.00002401
------------------------- = ----------------- = ----------- = 0.002951067
       [NH3]2eq                (0.0902)2          0.00813604

This is close to the original equilibrium constant value of 2.9 x 10-3.  The value of K calculated from these concentrations and rounded off to 2 significant figures is 3.0 x 10-3, which differs by only 1 digit in the last significant figure.  It must be remembered that a rounded value of X (expressed as 0.0049) was used to calculate the equilibrium concentrations.  The unrounded value is of X appears as 0.004861559 on a 10-digit scientific calculator.  If this is used, we get the following results:

[ [Ag(NH3)2]+ ]eq = [Cl-]eq = X = 0.004861559 mol/L

[NH3]eq = 0.10 - 2(0.004861559) = 0.10 - 0.009723118 = 0.090276882 mol/L

And plugging these unrounded values into the equilibrium constant expression,

{ [Ag(NH3)2]+ ]eq [Cl-]eq     (0.004861559)(0.004861559)    0.000023635
------------------------- = -------------------------- = ------------ = 0.0029
       [NH3]2eq                     (0.090276882)2         0.008149915

This time, we get "exactly" (within the precision limits of the 10-digit calculator) the original equilibrium constant value.  Equilibrium calculations tend to be somewhat susceptible to round-off error.  If you want to check the consistency of your calculated equilibrium concentrations, you may need to use more figures in your equilibrium concentrations than you would normally report as a final answer.

In the last 3 problems, we have calculated the molar solubility of AgCl in pure H2O, a 0.10 M NaCl solution, and a 0.10 M NH3 solution.  Let us now summarize and compare these results:

AGENT ROLE MOLAR SOLUBILITY OF AgCl EFFECT ON SOLUBILITY (COMPARED TO PURE H2O)
H2O Pure solvent 1.3 x 10-5 mol/L This IS pure H2O -- nothing to compare
0.10 M NaCl Provider of a common ion 1.8 x 10-9 mol/L About 7200 times less soluble
0.10 M NH3 Complexing agent 4.9 x 10-3 mol/L About 380 times more soluble

Notice that the NaCl suppresses the solubility (by a factor of about 7200) and the NH3 enhances it (by a factor of about 380).  Both of these effects can be explained through Le Chatelier's Principle.  The equilibrium of AgCl with its ions is represented by the equation

AgCl(s)   <---------->   Ag+(aq)    +   Cl-(aq)

The NaCl shifts this equilibrium to the left by supplying additional Cl- ions, which increases the rate of the reverse (right to left) reaction.  With a higher concentration of Cl-, not as much Ag+ is needed for the reverse reaction rate to equal the forward reaction rate.  Equilibrium will be established with a lower concentration of Ag+, and since AgCl is the sole source of Ag+ in an AgCl / NaCl system, this means a lower solubility of AgCl.

The NH3 shifts this equilibrium to the right by removing Ag+ ions from the solution (they are converted to [Ag(NH3)2]+ ions), which decreases the rate of the reverse (right to left) reaction.  With a lower concentration of Ag+, a higher concentration of Cl- is needed in order for the reverse reaction rate to equal the forward reaction rate.  As a result, equilibrium will be established with a greater concentration of Cl- in the solution.  Since AgCl is the sole source of Cl- in an AgCl / NH3 system, this means that the solubility of AgCl will be higher.

So far, we have looked at the water solubilities of two slightly soluble ionic compounds: BaCrO4 and AgCl.  For these two salts, we had the following equilibrium reactions and associated equilibrium constant (KSP) expressions:

BaCrO4(s)   <---------->   Ba2+(aq)   +   CrO42-(aq)          KSP = 1.2 x 10-10

AgCl(s)   <---------->   Ag+(aq)   +   Cl-(aq)                      KSP = 1.8 x 10-10

In each case, the salt produces  cations and anions in a one-to-one ratio.  Because these reactions have the same stoichiometry, their KSP values can be compared to determine their relative solubilities in water.  Since the KSP for AgCl is larger than that for BaCrO4, the AgCl must have a higher solubility in water.  In these notes, we found molar solubilities of 1.3 x 10-5 mol/L for AgCl and 1.1 x 10-5 mol/L for BaCrO4.  Their solubilities are  close, because their KSP values are close, but AgCl does have a higher molar solubility, as expected.

But now suppose we compare AgCl and Ag2CrO4.  Their equilibrium reactions and associated KSP values are:

AgCl(s)   <---------->   Ag+(aq)   +   Cl-(aq)                      KSP = 1.8 x 10-10

Ag2CrO4(s)   <---------->   2Ag+(aq)   +   CrO42-(aq)       KSP = 1.1 x 10-12

Ag2CrO4 has a smaller KSP, but does it have a lower solubility?  Read on and find out.  Shown below are the ICE table calculations for these two salts:

AgCl(s)   <---------->   Ag+(aq)   +   Cl-

 

[AgCl]

[Ag+]

[Cl-]

I

-----

0 0

C

-----

+X +X

E

-----

X X

KSP = [Ag+]eq [Cl-]eq = (X)(X) = X2 = 1.8 x 10-10

X = √ (1.8 x 10-10) = 1.3 x 10-5 mol/L = molar solubility

Ag2CrO4(s)   <---------->   2Ag+(aq)   +   CrO42-

 

[Ag2CrO4]

[Ag+]

[CrO42-]

I

-----

0 0

C

-----

+2X +X

E

-----

2X X

KSP = [Ag+]2eq [CrO42-]eq = (2X)2(X) = (4X2)(X) = 4X3 = 1.1 x 10-12

      1.1 x 10-12
X3 = ------------ = 2.75 x 10-13
          4

X = (2.75 x 10-13)1/3 = 6.5 x 10-5 mol/L = molar solubility

In the last equation above, the number is being raised to the 1/3 power, which is equivalent to taking the cube root.  Notice that even though Ag2CrO4 has a smaller KSP than AgCl, it has a larger molar solubility.  This is due to the difference in stoichiometry.  Since we end up with X3 in the equation for Ag2CrO4, X does not need to be as small to multiply out to the smaller value of KSP.  As this example shows, you can use KSP values to make direct comparisons of molar solubility only when the salts in question have the same stoichiometry.

A standard table of KSP values shows that numerous salts have a low solubility in water.  If an aqueous solution contains cations and anions that comprise a salt of low solubility, there exists the possibility that the solid form of the salt will precipitate from solution.  However, if the ion concentrations are low enough, the salt may remain in solution.  The way to find out what will happen is to do a QSP calculation.  As you may remember from the notes on "Chemical Equilibrium", you calculate a Q the same way you calculate a K except that the concentrations (or pressures of gases, if using pressure-based equilibrium constants) don't necessarily have to be those of an equilibrium mixture.  The value of Q is compared to that of K, and the following interpretations are made:

Determination of Equilibrium Status Based on Q Values

Comparison of Q to K

Interpretation Based on This Comparion

Q = K

System is at equilibrium

Q < K

Reaction will occur to the right (conversion of reactant to products)

Q > K

Reaction will occur to the left (conversion of products to reactants)

To decide if a salt will precipitate from solution, the ion concentrations are multiplied together (raised to powers, if appropriate) to produce QSP.  If QSP is less than KSP, this indicates the system is not at equilibrium and reaction would need to proceed to the right.  Since KSP values are given for the reaction written with the solid on the left and the ions on the right, reaction to the right means more solid could dissolve, even in the presence of the existing ion concentrations.  No precipitation will occur when QSP < KSP.  But if QSP > KSP, the system is not at equilibrium and needs to react to the left to reach equilibrium.  This means the ion concentrations are too high, and some precipitation will occur until the ion concentrations are lowered to their equilibrium concentrations.  The solid appears on the left side of the equation, so saying the reaction will occur to the left means ions in solution will be converted to the solid salt.

Example Problem: Suppose you mix 22.0 mL of 0.100 M AgNO3 with 45.0 mL of 0.0260 M NaC2H3O2. The two possible products of the metathesis reaction are NaNO3 and AgC2H3O2.  NaNO3 is a freely soluble salt, but AgC2H3O2 has limited solubility:

AgC2H3O2(s)    <---------->    Ag+(aq)    +    C2H3O2-(aq)       Ksp = 1.9 x 10-3

Would a precipitate of AgC2H3O2 form when these solutions are mixed?

Discussion and Answers

We need to find the ion concentrations upon mixing the solutions, so we can use them in the QSP calculation.  The number of moles of Ag+ contributed by the 0.100 M AgNO3 is calculated as follows:

            1 x 10-3 L     0.100 mol
22.0 mL  x  ----------  x  ---------  =  2.20 x 10-3 mol
              1 mL           1 L

The number of moles of C2H3O2- contributed by the 0.0260 M NaC2H3O2 is calculated as follows:

            1 x 10-3 L     0.0260 mol
45.0 mL  x  ----------  x  ----------  =  1.17 x 10-3 mol
              1 mL            1 L

If we assume the volumes are additive, then mixing 22.0 mL of one solution with 45.0 mL of another solution will give a combined volume of 67.0 mL, which is 0.0670 L.  Now the ion concentrations can be calculated:

        2.20 x 10-3 mol
[Ag+] = ---------------  =  3.28 x 10-2 mol/L
          0.0670 L

            1.17 x 10-3 mol
[C2H3O2-] = ----------------  =  1.75 x 10-2 mol/L
             0.0670 L

Based on the equilibrium reaction:

AgC2H3O2(s)    <---------->    Ag+(aq)    +    C2H3O2-(aq)

the QSP expression is

QSP = [Ag+] [C2H3O2-] = (3.28 x 10-2)(1.75 x 10-2) = 5.74 x 10-4

Our value for QSP is 5.74 x 10-4 but KSP is 1.9 x 10-3.  So QSP < KSP.  The system is not at equilibrium, and reaction would need to go to the right.  This is in the direction of ions in solution -- that is, the dissolved salt.  Therefore, the solution is not saturated, and no precipitation will occur.  In fact, we could dissolve additional AgC2H3O2 in this solution, even in the presence of the Ag+ and C2H3O2- ions that are already there.

The final topic to be addressed in these notes is fractional precipitation.  This refers to the precipitation of one ion from solution while leaving another ion in solution.  When such an effect is possible, it allows ions to be separated from each other.  Such separations are an important part of qualitative analysis labs.  A solution containing a mixture of several ions is processed in such a way as to take advantage of the differing solubilities of various salts containing those ions.  The goal is to get every ion in a separate test tube so it can then be uniquely identified, without interference from other ions in the mixture.  In order to achieve good separation of two ions from each other, the salts they form with a third ion must have substantially different solubilities.  This allows nearly 100% of one ion to be removed from solution before the other ion begins to precipitate.  If the two salts in question have KSP values that are close to each other, a good separation can not be achieved.  The following problems pertain to fractional precipitation and its use in separating ions from each other.

Example Problem:  Lead(II) chloride, PbCl2 and lead(II) iodide, PbI2 are both salts with limited solubility in water. The chemical equations for their equilibria in distilled water are as follows:

PbCl2(s)    <---------->    Pb2+(aq)    +    2Cl-(aq)       Ksp = 1.7 x 10-5

PbI2(s)    <---------->    Pb2+(aq)    +    2I-(aq)       Ksp = 9.8 x 10-9

If a solution contained Cl- and I- ions at a concentration of 0.100 M each, and you began adding Pb2+ ions to the solution, which substance would be the first to precipitate out of the solution?

Discussion and Answers

This is a very simple problem.  Since the two equilibrium equations have the same stoichiometry, their KSP values can be directly compared.  PbI2 has a substantially smaller KSP than PbCl2.  Therefore, PbI2 has a lower molar solubility, and since the Cl- and I- ions have the same starting concentration, PbI2 will be the first to precipitate when Pb2+ ions are added to the solution.

Example Problem:  For the anion that precipitated first in the previous problem, what concentration would remain in solution (in units of moles per liter) at the moment the second ion began precipitating? (Assume no change in solution volume as the Pb2+ ions are added.)

Discussion and Answers

To solve this problem, we first calculate the concentration of Pb2+ needed to begin precipitation of PbCl2 from solution.  Then we can calculate what concentration of I- would be in equilibrium with Pb2+ at that concentration.

PbCl2 first begins to precipitate when the ions can be in equilibrium with the solid salt.  Based on the stoichiometry of its ionization,

PbCl2(s)   <---------->   Pb2+(aq)   +   2Cl-(aq)

we know that

KSP = [Pb2+]eq [Cl-]2eq

Solving this for the concentration of lead(II) ion, we get

            KSP          1.7 x 10-5
[Pb2+]eq = ---------  =  ----------  =  1.7 x 10-3 mol/L
           [Cl-]2eq       (0.100)2

Precipitation of PbCl2 will begin when the Pb2+ concentration reaches 0.0017 M.  Any Pb2+ concentration larger than this would produce a QSP value that is greater than KSP.  As we have seen, if Q > K, reaction will take place to the left, which is the solid side of the equilibrium equation for slightly soluble salts.

Now we can address the question of what concentration of I- remains in solution at the moment the Cl- begins precipitating from solution.  Since the PbI2 has a substantially smaller KSP than PbCl2, the PbI2 begins precipitating first and most of the I- is removed from solution by the time the Cl- begins to precipitate.

Based on the stoichiometry of the ionization of PbI2 in water,

PbI2(s)   <---------->   Pb2+(aq)   +   2I-(aq)

we know that

KSP = [Pb2+]eq [I-]2eq

Solving this for the concentration of iodide ion squared gives

           KSP             9.8 x 10-9
[I-]2eq = ---------  =  -----------  =  5.8 x 10-6
          [Pb2+]eq      1.7 x 10-3 

In the above equation, the value 1.7 x 10-3 is the concentration of Pb2+ needed to begin precipitation of PbCl2.  The number obtained from this calculation is the square of the concentration, not the concentration itself, so now we need to take the square root:

[I-]eq = √ (5.8 x 10-6)  =  2.4 x 10-3 mol/L

The I- concentration started out at 0.100 mol/L, and by the time the Cl- began precipitating, the I- concentration had been reduced to 2.4 x 10-3 mol/L.

Example Problem:  What percentage of the I- was removed from solution in the previous problem before the Cl- began precipitating along with it?

Discussion and Answers

In order not to precipitate Cl- ions, we were not able to let the Pb2+ concentration exceed 1.7 x 10-3 mol/L. That is when the PbCl2 just begins to precipitate.  So that concentration of Pb2+ offers the maximum separation of I- and Cl-.  We saw previously that when the Pb2+ concentration is 1.7 x 10-3 mol/L, the concentration of I- remaining in solution is 2.4 x 10-3 mol/L. The starting concentration was 0.100 mol/L.  This means the percentage of I- remaining in solution is

            2.4 x 10-3 mol/L
100%   x   -----------------  =  2.4%
              0.100 mol/L

So 2.4% of the original I- remains in solution, and can't be removed without precipitating some of the Cl- along with it.  The percentage of I- that could be removed without contamination from Cl- is therefore

100.0% - 2.4%  =  97.6%

In summary, we have seen that salts of low solubility establish an equilibrium between their solid form and their ions in aqueous solution.  The equilibrium constant for the reaction, referred to as KSP, indicates to what extent the salt will dissolve.  In general, the smaller the KSP value is, the less soluble the salt.  However, we can make direct comparisons only when the salts we wish to compare have the same stoichiometry for their ionization in water.  We have also seen ways that the solubility of a slightly soluble salt may be modified.  Slightly soluble salts become even less soluble if one of the ions found in the salt is already present in solution from another source.  We call this the common ion effect.  Finally, we can enhance the solubility of a slightly soluble salt by adding a complexing agent to the solution.  The complexing agent binds to one of the ions from the salt, to form a complex ion.  Both effects -- the reduction of solubility seen in the presence of a common ion, and the enhancement in solubility seen in the presence of a complexing agent -- can be explained on the basis of Le Chatelier's Principle.  The presence of a common ion means a higher ion concentration, and this shifts the equilibrium toward the solid salt, meaning a lower solubility will be observed.  A complexing agent removes one of the salt's ion from solution, and this decrease in ion concentration shifts the equilibrium toward the ionized form of the salt.  Therefore, the solubility will be higher in the presence of the complexing agent.  In solving these equilibrium problems, we have seen that the mathematics can sometimes become tedious, but we have also been fortunate to discover that in many of those cases, approximations can be made to greatly simplify the problem.  In these notes, for example, we have been able to avoid having to solve the quadratic formula by taking advantage of appropriate approximations.

This concludes this set of general chemistry notes.  I hope you have found them to be helpful.  If you have any questions about the content of these notes, or any suggestions on how they can be made more helpful, please let me know.