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810 Chapter 15 | Equilibria of Other Reaction Classes
Example 15.4
Determination of Molar Solubility from Ksp, Part II
The Ksp of calcium hydroxide, Ca(OH)2, is 1.3 10–6. Calculate the molar solubility of calcium hydroxide. Solution
The solubility product of calcium hydroxide is 1.3 10–6.
The reaction is:
First, write out the solubility product expression:
Create an ICE table, leaving the Ca(OH)2 column empty as it is a solid and does not contribute to the Ksp:
At equilibrium:
Therefore, the molar solubility of Ca(OH)2 is 6.9 10–3 M.
Check Your Learning
The Ksp of PbI2 is 1.4 10–8. Calculate the molar solubility of lead(II) iodide.
Answer: 1.5 10–3 M
Note that solubility is not always given as a molar value. When the solubility of a compound is given in some unit other than moles per liter, we must convert the solubility into moles per liter (i.e., molarity) in order to use it in the solubility product expression. Example 15.5 shows how to perform those unit conversions before determining the solubility product equilibrium.
Example 15.5
Determination of Ksp from Gram Solubility
Many of the pigments used by artists in oil-based paints (Figure 15.3) are sparingly soluble in water. For example, the solubility of the artist’s pigment chrome yellow, PbCrO4, is 4.6 10–6 g/L. Determine the solubility product for PbCrO4.
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