**K _{sp}** stands for the solubility product constant for an equilibrium expression. different ionic solids have different K

_{sp}values. The K

_{sp}can be calculated as ([concentration of product1]*[concentration of product 2])/[concentration of reactant]. However, when calculating the K

_{sp}, since we do not include any solids or liquids in the expression, and K

_{sp}is only used with solid compounds, we do not include the solid reactant in the expression. Thus we are left with the final equation of K

_{sp}= [concentration of product 1]*[concentration of product 2]. Also when calculating K

_{sp}the coefficients in the reaction are correlated to the exponents in the expression. In other words, the exponents in the expression are the coefficients in the equation.

For example, suppose 2Na^{+} (aq) + SO_{4}^{2-}(aq) —> Na_{2}SO_{4} (aq), then the equilibrium expression would be K_{sp} = [Na_{2}SO_{4}]/ [Na^{+}]^{2} [SO_{4}^{2-}].

Since Na^{+} has a coefficient of 2 in the equation, the exponent on Na^{+} is 2 in the expression.

By using the K_{sp,} we can determine how soluble a substance is. There are no units for K_{sp.} It is also important to note that K_{sp} and solubility are two different things. K_{sp} is one value, it doesn’t change or vary. Solubility is a position, meaning its values can vary and does not have a set value. One can calculate the K_{sp} from solubility and vice versa.

For example: Cu_{2}S_{3} (s) 3S^{2-}(aq) + 2Cu^{3+}(aq). The solubility for Cu_{2}S_{3} is 1.0 X 10^{-5}M. Calculate the K_{sp.}

K_{sp} = [S^{2-}]^{3}[Cu^{3+}]^{2} because K_{sp} doesn’t include solids, you omit Cu_{2}S_{3} from the expression. Because the coefficients are 3 and 2, the exponents are 3 and 2.

Because the mole ratio for Cu_{2}S_{3} and S^{2- } is 1-3 the solubility for S^{2-}is 1.0 X 10^{-5} M Cu_{2}S_{3 X } 3mol S^{2-}/ 1 mol Cu_{2}S_{3} = 3.0 X 10^{-5}M S^{2-}. Because the mole ratio for Cu_{2}S_{3} is 1-2 the solubility for Cu^{3+} is 1.0 X 10^{-5} M Cu_{2}S_{3 X } 2mol Cu^{3+}/ 1 mol Cu_{2}S_{3} = 2.0 X 10^{-5}M Cu^{3+}.

next you plug the values into the K_{sp} expression.

K_{sp}=[3.0 X 10^{-5}M]^{3}[2.0 X 10^{-5}M]^{2}= 1.08 X 10^{-3} units are not needed.

One can also calculate solubility from K_{sp}

For example: Sr(OH)_{2}(s)—> 2OH^{–}(aq) + Sr^{2+}(aq). If the K_{sp} is 4.0 X 10^{-6} what is the solubility of OH^{–}

First you write the expression. K_{sp =} [OH^{–}]^{2}[Sr^{2+}]

then you make an I.C.E. table.

Sr(OH)_{2}(s)—> 2OH^{–}(aq) + Sr^{2+}(aq)

I. 0 0 For this problem, we will omit

C. +2x +x the values for Sr(OH)_{2} because it

E 2x x is not included in the expression.

Next we plug it into the expression: 4.0 X 10^{-6}= [2x]^{2}[x]. We we solve for x, we find that X = 1.0 X 10^{-8}M. because there is a 2 in front of OH^{–} we multiply it by 2. Thus the solubility of OH- is 2.0 X 10^{-8}M.

Comments are closed