http://www.ck12.org Chapter 19. Equilibrium
Sample Problem 19.4: The Common Ion Effect
What is the concentration of zinc ion in 1.00 L of a saturated solution of zinc hydroxide to which 0.040 mol of NaOH
has been added?
Step 1: List the known quantities and plan the problem.
Known
- Ks p= 3.0× 10 −^16 (Table19.5)
- moles of added NaOH = 0.040 mol
- volume of solution = 1.00 L
Unknown
- [Zn^2 +] =? M
Express the concentrations of the two ions relative to the variable s. The concentration of the zinc ion will be equal
to s, while the concentration of the hydroxide ion will be equal to 0.040 + 2s. Note that the value of s in this case is
not equal to the value of s when zinc hydroxide is dissolved in pure water.
Step 2: Solve.
The Ks pexpression can be written in terms of the variable s.
Ks p= [Zn^2 +][OH−]^2 = (s)(0.040+2s)^2Because the value of the Ks pis so small, we can make the assumption that the value of s will be very small compared
to 0.040. This simplifies the mathematics involved in solving for s.
Ks p= (s)(0.040)^2 = 0.0016s = 3.0× 10 −^16
s=Ksp
[OH−]^2=
3. 0 × 10 −^16
0. 0016
= 1. 9 × 10 −^13 M
The concentration of the zinc ion is equal to s, so [Zn^2 +] = 1.9× 10 −^13 M.
Step 3: Think about your result.
The relatively high concentration of the common ion, OH−, results in a very low concentration of the zinc ion. The
molar solubility of the zinc hydroxide is less in the presence of the common ion than it would be in pure water.
Practice Problem- Determine the concentration of silver ions in 1.00 L of a saturated solution of silver chloride to which 0.0020
mol of sodium chloride has been added.
Lesson Summary
- In a saturated solution, an equilibrium exists between the dissolved and undissolved solute, at which point the
rate of dissolution is equal to the rate of recrystallization. The equilibrium constant expression for this type of
equilibrium is called a solubility product constant (Ks p).