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Common Ions in Heterogeneous Equilibria
Skills to develop
 Recognize common ions from various salts, acids, and bases.
 Calculate concentrations involving common ions.
 Calculate ion concentrations involving chemical equilibrium.
 Apply the Newton's method to solve cubic equations.
Common Ions in Heterogeneous Equilibria
The solubility products K_{sp}s 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^{}
CaCl_{2} = Ca^{2+} + 2 Cl^{}
AlCl_{3} = Al^{3+} + 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^{}], [Ca^{2+}], and [H^{+}] in a solution containing
0.10 M each of NaCl, CaCl_{2}, and HCl.
Solution
Apparently, due to the conservation of ions, we have
[Na^{+}] = [Ca^{2+}] = [H^{+}] = 0.10 M.
but
[Cl^{}] = 0.10 (due to NaCl)
+ 0.20 (due to CaCl_{2})
+ 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? K_{sp} for AgCl is 1.0E10.
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 K_{sp}, we have
x (0.05 + x) = K_{sp};
Thus,
x = 1.0e10 / 0.05 = 2e9 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 Li_{2}CO_{3} in a solution
containing also 0.10 M of K_{2}CO_{3}?
The solubility product for Li_{2}CO_{3} is 0.0017.
Solution
In this case, CO_{3}^{2} is the common ion between the two salts.
K_{2}CO_{3} = 2 K^{+} + CO_{3}^{2}
Li_{2}CO_{3} = 2 Li^{+} + CO_{3}^{2}
This question implies that K_{2}CO_{3} is soluble.
Thus, the only heterogeneous equilibrium to be considered is
K_{2}CO_{3} = 2 K^{+} + CO_{3}^{2}
Let [Li^{+}] = 2 x, then [CO_{3}^{2}] = 0.10 + x.
Thus, we can write down the equilibrium equation again, and write
the concentrations below the chemical formula:
Li_{2}CO_{3} = 2 Li^{+} + CO_{3}^{2}
2 x 0.10+x
[Li^{+}]^{2} [CO_{3}^{2}] = (2 x)^{2} (0.10+x) = 0.0017
Expanding the formula results in
x^{3} + 0.10 x^{2}  0.00043 = 0.
A general method to solve this equation is to use the Newton's method.
For which, we assume that
y = x^{3} + 0.10 x^{2}  0.00043 = 0.
We assume the value of x = to 0.06, and substitute in the equation
to calculate y.
y = 0.06^{3} + 0.10*0.06^{2}  0.00043 = 0.000146
We now assume x = 0.05, and substitute in the equation to evaluate y,
y = 0.05^{3} + 0.10*0.05^{2}  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 = 0.053^{3} + 0.10*0.053^{2}  0.00043 = 2.2e7
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 = 0.0531^{3} + 0.10*0.0531^{2}  0.00043 = 1.7e6
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 Li_{2}CO_{3} in a solution
containing also 0.10 M of K_{2}CO_{3} 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 K_{sp} 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 Li_{2}CO_{3} in a solution
which is 0.20 M in Na_{2}CO_{3}. The solubility product
of LI_{2}CO_{3} is 1.7e3 M^{3}.
Solution
Since the solution contain 0.20 M Na_{2}CO_{3},
[CO_{3}^{2}] = 0.20 M. Assume the solubility to be
x M of Li_{2}CO_{3}, then we have
Li_{2}CO_{3} = 2 Li^{+} + CO_{3}^{2}
2 x x + 0.20 M
y  x  Remarks


2e4  0.05  any small value for x

5e4  0.06  y larger, try x < 0.05

5e5  0.04  0.04 < x < 0.05

8e6  0.0425  0.04 < x < 0.0425

1e5  0.0415  0.0415 < x < 0.0425

3e6  0.0420  0.0415 < x < 0.0420

5e6  0.0419  0.0415 < x < 0.0419

Thus,
(2 x)^{2} (x+0.20) = 1.7e3 M^{3}
4x^{3} + 0.20 x^{2} = 1.7e3 M^{3}
x^{3} + 0.20 x^{2}  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 = x^{3} + 0.20 x^{2}  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.
Confidence Building Questions

The solubility product K_{sp} for bismuth sulfide
Bi_{2}S_{3} is 1.6E72 at 25 deg C. What is the molar
solubility of bismuth sulfide in a solution that is 0.0010 M in
sodium sulfide Na_{2}S?
Answer 1.08e13
Consider...
x = molar solubility; (2x)^{2} (0.0010 + 3x)^{3} = K_{sp}; x = ?

John poured 1.0 mL of 0.10 M NaCl, 1.0 mL of 0.10 M KOH, and and 1.0 mL
0.20 HCl solutions together and then he made the total volume to be 100.0 mL.
What is the [Cl^{}] in the final solution?
Answer 0.0030 M
Consider...
Evaluate [Na^{+}], [K^{+}], [H^{+}] and
[OH^{}] for fun!

The K_{sp} for AgCl is 1.0E10.
From which of the following solutions would silver chloride precipitate?
A: A solution 0.10 M in Ag^{+} and 1.00 M in Cl^{}
B: A solution 1.0e5 M in Ag^{+} and 0.20 M in Cl^{}
C: A solution 1.0E7 M in Ag^{+} and 1.0E7 M in Cl^{}
a. A only
b. B only
c. C only
d. A and B
e. A, B, and C
Answer d
Consider...
Calculate the [Ag^{+}] [Cl^{}] for A, B, and C, and compare
their values with K_{sp}.
Substance  K_{sp}


magnesium hydroxide  1.2E11

magnesium carbonate  1.6E5

magnesium fluoride  6.4e9

 Addition of which of the following substances will cause the
precipitation of a salt from one liter of a 1e4 M Mg^{2+} solution?
a. 1E4 mole NaOH
b. 1E1 mole nitric acid
c. 1E5 mole potassium acetate
d. 1E4 mole ammonium nitrate
e. 1E2 mole sodium fluoride
Answer e
Consider...
A lot of calculations to figure out, but it's a common ion problem.

The K_{sp} for strontium chromate is 3.6E5 and K_{sp} for barium chromate
is 1.2E10. What concentration of potassium chromate will
precipitate the maximum amount of either the barium or the strontium
chromate from an equimolar 0.10 M solution of barium and strontium
ions without precipitating the other?
Answer 3.6e4
Consider...
This problem requires some thinking.

Iron(II) hydroxide is only sparingly soluble in water at 25 C; its
K_{sp} is equal to 7.9E16. Calculate the solubility of
iron(II) hydroxide in a solution of pH 6.0.
Answer 7.9 M
Consider...
[OH^{}] = 10^{(14+6)} = 1e8. x * (1e8 + x)^{2} = K_{sp}; x = ?
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