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The Solubility Product Constant

Let's say we are adding silver chloride (AgCl) into a liter of water. We can express the equilibrium as follows:

image\24868 ch 8.png

The solubilityproduct (Ksp) is defined as the equilibrium constant for the solubility equilibrium of ionic compounds. The solubility product of AgCl is expressed as:

image\24877 ch 8.png

The solubility product constant is equal to the product of the concentrations of the ions in a saturated solution, each concentration raised to a power equal to its coefficient (number of moles of individual ions formed per mole of the compound) Take a look at the solubility product equation of calcium fluoride (CaF2).

image\24886 ch 8.png

It is imperative that you understand these ideas involving the solubility product constant. Let's look at another example. One mole of aluminum hydroxide [Al(OH)3] gives one mole of aluminum ions and three moles of hydroxide ions in solution.

image\Ch 8 sec F, g4.png

From the balanced equation, we can see that the coefficient of Al3+ is 1 and that of OH is 3. So, the solubility constant (Ksp) expression for Al(OH)3 is given by:

Ksp = [Al3+] [OH]3

The solubility product depends on the temperature, and at a given temperature it is constant for a particular ionic compound. Molar solubility is defined as the number of moles of solute dissolved in one liter of its saturated solution. Using the molar solubility of a compound, the solubility product of that compound can be determined or vice versa.

Example 8-2

Calculate the molar solubility of silver chloride (AgCl). The solubility product (Ksp) of AgCl is 1.8 x 10–10.



image\24895 ch 8.png

Ksp= [Ag+] [Cl]


Let x be the molar solubility of AgCl. So we can write:


image\24905 ch 8.png

Ksp= [Ag+] [Cl] = (x) (x) = (x)2 = 1.8 x 10–10




Example 8-3

If 'x' represents the molar solubility of Al(OH)3, find the expression for the solubility product of Al(OH)3 in terms of x. What is the molar solubility of Al(OH)3, if the solubility product constant is 2 x 10–33?



image\Ch 8 Example 8-3.png

image\Ch 8 Example 8-3b.png


You can approximate numbers to find the correct answer among the choices in the MCAT.
In this example, to the 4th root of 7.4 x 10–35, first we can rewrite this as 74 x 10–36. The 4th root of 81 is 3. Since 74 is less than 81, we can approximate the 4th root of 74 close to 2.8 or 2.9. If you had trouble realizing that, think of the 4th root of 16 which is 2. Again, the best approximation for the 4th root of 74 is a number below and close to 3. The 4th root of 10–36 is 10–9. Such approximation-techniques can save time during the test.


Ion Product

Ion product is the product of the concentrations of the ions from the compound (solute) in a solution, each concentration raised to a power equal to its coefficient in the balanced equation. In other words, the expression for the ion product is the same as that of Ksp. In a saturated solution, the ion product is equal to Ksp. If the ion product is greater than Ksp, the solution is supersaturated and can undergo precipitation. If the ion product is less than Ksp, the solution is unsaturated. Thus, by comparing the ion product of a compound against its Ksp, we can predict whether or not precipitation is likely to occur.

image\Ch 8 Ion Product table.png

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