Chesapeake Campus – Chemistry 112 Laboratory
LAB #10 –The Common Ion Effect
Objectives
After completing this lab you will be able to:
- Calculate the [OH–] using a pH probe.
- Calculate the molar solubility of Ca(OH)2 and Mg(OH)2.
- Calculate the solubility constants of Ca(OH)2 and Mg(OH)2.
Introduction
Solubility
In CHM 111 you were introduced to the idea that some ionic compounds are soluble while others are insoluble. There are many salts that are classified as “slightly soluble” or “marginally soluble.” Slightly soluble salts are in a constant equilibrium in solution. They typically have low K values indicating they are reactant favored. For example, lead II bromide ionizes in solution as
where the equilibrium expression is depicted as:
Notice that the equilibrium expression does not include the lead II bromide as this is a solid. The equilibrium constant for a slightly soluble salt is given as Ksp or the solubility constant.
The molar solubility of a compound is the amount of that compound that can be dissolved to make a saturated solution. Lead II bromide has a molar solubility of 4.4 x 10-2 . This means that only 0.044 mol of lead II bromide can be dissolved in a 1 L solution. According to the equation above, we can tell that every lead II bromide that dissolves produces 1 lead ion and 2 bromide ions. Therefore in solution we can determine the concentration of both ions as:
This indicates the [Pb2+] concentration is 4.4 x 10-2 M and the bromide ion concentration is 8.8 x 10-2 M.
This can be plugged into the Ksp equation to find the solubility constant:
LeChâtelier’s Principle
LeChâtelier’s Principle states that the equilibrium will shift to alleviate the stress on a system. Therefore, if NaBr was added to the lead II bromide solution above, it would add in Br–. This addition of “product” ion would cause the system to shift to the left to produce more solid lead II bromide. Since an ion common to the original salt was added, this is often called the common ion effect. The shift to the left in equilibrium is accompanied by a reduction in the amount of salt that remains in solution. Thus the molar solubility is decreased with the addition of a common ion.
Determining the Molar Solubility Using pH
In today’s lab, you will be calculating the molar solubility of two slightly soluble hydroxide salts; Ca(OH)2 and Mg(OH)2. Calcium hydroxide ionizes according to the following equation:
Equation 1
from this equation we can see that
Equation 2
The hydroxide ion concentration will be determined by taking a pH reading for the solution. The pH allows the pOH to be calculated using Equation 3.
Equation 3
The hydroxide ion concentration can then be calculated by solving the pOH equation for [OH=].
Equation 4
Equation 5
Once the hydroxide ion concentration is calculated, it is possible to use Equation 2 to determine the concentration of calcium in solution and the molar solubility of calcium hydroxide. The concentrations of calcium and hydroxide ions can be plugged into the Ksp equation for calcium hydroxide:
Equation 6
Equation 6 can also be rewritten with the calcium ion concentration substituted according to Equation 2 to give Equation 7 below.
Equation 7