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Common Ions in Heterogeneous Equilibria

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Common Ions in Heterogeneous Equilibria

The solubility products Ksps are equilibrium constants in hetergeneous equilibria. If several salts are present in a system, they all ionize in the solution. If the salts contain a common cation or anion, these salts contribute to the concentration of the common ion. Contributions from all salts must be included in the calculation of concentration of the common ion.

For example, a solution containing sodium chloride and potasium chloride will have the following relationship:

[Na+] + [K+] = [Cl-] Consideration of charge balance or mass balance or both leads to the same conclusion.

Common Ions

When you dissolve NaCl and KCl in a solution, the Cl- ions are common to both salts. In a system containing NaCl and KCl, the Cl- ions are Common ions. NaCl = Na+ + Cl-
KCl = K+ + Cl-
CaCl2 = Ca2+ + 2 Cl-
AlCl3 = Al3+ + 3 Cl-
AgCl = Ag+ + Cl- The following example show how to the concentration of the common ion is calculated. Confidence Building Questions on this page help you practice problem solving involving common ions.

When you dissolve AgCl in a solution already containing NaCl (actually Na+ and Cl- ions), the Cl- ions come from the ionization of both AgCl and NaCl. Thus, [Cl-] differs from [Ag+].

Example 1

What are [Na+], [Cl-], [Ca2+], and [H+] in a solution containing 0.10 M each of NaCl, CaCl2, and HCl.

Solution
Apparently, due to the conservation of ions, we have

[Na+] = [Ca2+] = [H+] = 0.10 M. but 
    [Cl-] =   0.10 (due to NaCl)
            + 0.20 (due to CaCl2)
            + 0.10 (due to HCl)
         = 0.40 M

Discussion:
John poured 10.0 mL of 0.10 M NaCl, 10.0 mL of 0.10 M KOH, and 5.0 mL 0.20 HCl solutions together and then he made the total volume to be 100.0 mL. What is [Cl-] in the final solution?
[Cl-] = (0.1 M*10 mL+0.2 M*5.0 mL)/100.0 mL = 0.020 M

Example 2

What is the solubility (in moles per liter) of AgCl in a solution that is 0.05 M in KCl? Ksp for AgCl is 1.0E-10.

Solution
Assume x = [Ag+] to be the molar solubility, then we can write the equilibrium concentration below the formula of the equilibrium. 

 AgCl -> Ag+ + Cl-
         x    0.05+x   <-- equilibrium concentration
By definition of the Ksp, we have x (0.05 + x) = Ksp; Thus, x = 1.0e-10 / 0.05 = 2e-9 M Note that 0.05 + x is approximately 0.05, and you may use this approximation for the calculation.

Example 3

What is the solubility of Li2CO3 in a solution containing also 0.10 M of K2CO3? The solubility product for Li2CO3 is 0.0017.

Solution
In this case, CO32- is the common ion between the two salts.

K2CO3 = 2 K+ + CO32-
Li2CO3 = 2 Li+ + CO32- This question implies that K2CO3 is soluble. Thus, the only heterogeneous equilibrium to be considered is K2CO3 = 2 K+ + CO32- Let [Li+] = 2 x, then [CO32-] = 0.10 + x. Thus, we can write down the equilibrium equation again, and write the concentrations below the chemical formula: 
    Li2CO3 = 2 Li+ + CO32-
             2 x    0.10+x
[Li+]2 [CO32-] = (2 x)2 (0.10+x) = 0.0017
Expanding the formula results in x3 + 0.10 x2 - 0.00043 = 0. A general method to solve this equation is to use the Newton's method. For which, we assume that y = x3 + 0.10 x2 - 0.00043 = 0. We assume the value of x = to 0.06, and substitute in the equation to calculate y. y = 0.063 + 0.10*0.062 - 0.00043 = 0.000146 We now assume x = 0.05, and substitute in the equation to evaluate y, y = 0.053 + 0.10*0.052 - 0.00043 = -0.000055 Since the values for y calculated for x = 0.05 and 0.06 have different sign, the x value should lie between 0.05 and 0.06. We further assume the value for x = 0.053, we obtain a y value y = 0.0533 + 0.10*0.0532 - 0.00043 = -2.2e-7 Thus, the value of x should be greater than 0.053. We can increase x to 0.0531 or 0.0532, and evaluate y again: y = 0.05313 + 0.10*0.05312 - 0.00043 = 1.7e-6 Thus, x value should lie between 0.0531 and 0.0530. By this method, we have evaluated x values to three significant figures. We only need two significant figures due to the nature of the data in the problem. Thus we have [Li+] = 2 x = 0.106 M

Discussion
The molar solubility of Li2CO3 in a solution containing also 0.10 M of K2CO3 is 0.053 mole per liter, but [Li+] = 2 x = 0.106 M.

This example illustrates the Newton's method for solving cubic equations. Note that if the Ksp is small, then x is a very small value. In this case, 0.10 + x is approximately 0.10. You do not need to use the Newton's method in this case.

Example 4

What is the solubility of Li2CO3 in a solution which is 0.20 M in Na2CO3. The solubility product of LI2CO3 is 1.7e-3 M3.

Solution
Since the solution contain 0.20 M Na2CO3, [CO32-] = 0.20 M. Assume the solubility to be x M of Li2CO3, then we have

Li2CO3 = 2 Li+ + CO32-
                2 x       x + 0.20 M
y x Remarks
2e-4 0.05 any small value for x
5e-4 0.06 y larger, try x < 0.05
-5e-5 0.04 0.04 < x < 0.05
8e-6 0.0425 0.04 < x < 0.0425
-1e-5 0.0415 0.0415 < x < 0.0425
3e-6 0.0420 0.0415 < x < 0.0420
-5e-6 0.0419 0.0415 < x < 0.0419
Thus,
(2 x)2 (x+0.20) = 1.7e-3 M3
4x3 + 0.20 x2 = 1.7e-3 M3
x3 + 0.20 x2 - 0.00043 = 0
There is no definite way to solve this equation, but the Newton's method is very useful. In this method, we let
y = x3 + 0.20 x2 - 0.00043
We guesstimate x and evaluate y using the above expression. The solution is a value for x so that y=0. We refine x values between two x values that give positive and negative y values in progression as shown in the Table on the right.

Discussion
This is the same type of question as Example 3. As the carbonate concentration increases from 0.10 to 0.20 M, the solubility of lithium carbonate reduces to 0.042 M from 0.053 M. Note that the lithium ion concentration is 0.084 M in this case.

Note that when used to treat depression, lithium carbonate is usually called lithium in the health care profession.

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