# The Solubility Product Constant

**solubilityproduct**(

*K*) is defined as the equilibrium constant for the solubility equilibrium of ionic compounds. The solubility product of AgCl is expressed as:

_{sp}

_{2}).

_{3}] gives one mole of aluminum ions and three moles of hydroxide ions in solution.

^{3+}is 1 and that of OH

^{â€“}is 3. So, the solubility constant (

*K*sp) expression for Al(OH)

_{3}is given by:

*K*sp = [Al^{3+}] [OH^{â€“}]^{3}

*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.

Calculate the molar solubility of silver chloride (AgCl). The solubility product (*K _{sp}*) of AgCl is 1.8 x 10

^{â€“10}.

*K _{sp}*= [Ag

^{+}] [Cl

^{â€“}]

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

*K _{sp}*= [Ag

^{+}] [Cl

^{â€“}] = (

*x*) (

*x*) = (

*x*)

^{2}= 1.8 x 10

^{â€“10}

_{}

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^{â€“33}?

You can approximate numbers to find the correct answer among the choices in the MCAT.

In this example, to the 4^{th} root of 7.4 x 10^{â€“35}, first we can rewrite this as 74 x 10^{â€“36}. The 4^{th} root of 81 is 3. Since 74 is less than 81, we can approximate the 4^{th} root of 74 close to 2.8 or 2.9. If you had trouble realizing that, think of the 4^{th} root of 16 which is 2. Again, the best approximation for the 4^{th} root of 74 is a number below and close to 3. The 4^{th} root of 10^{â€“36} is 10^{â€“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

*K*

_{sp}. In a saturated solution, the ion product is equal to

*K*

_{sp}. If the ion product is greater than

*K*

_{sp}, the solution is supersaturated and can undergo precipitation. If the ion product is less than

*K*

_{sp}, the solution is unsaturated. Thus, by comparing the ion product of a compound against its

*K*

_{sp}, we can predict whether or not precipitation is likely to occur.