Ksp stands for the solubility product constant for an equilibrium expression. different ionic solids have different Ksp values. The Ksp can be calculated as ([concentration of product1]*[concentration of product 2])/[concentration of reactant]. However, when calculating the Ksp, since we do not include any solids or liquids in the expression, and Ksp 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 Ksp = [concentration of product 1]*[concentration of product 2]. Also when calculating Kspthe 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) + SO42-(aq) —> Na2SO4 (aq), then the equilibrium expression would be Ksp = [Na2SO4]/ [Na+]2 [SO42-].
Since Na+ has a coefficient of 2 in the equation, the exponent on Na+ is 2 in the expression.
By using the Ksp, we can determine how soluble a substance is. There are no units for Ksp. It is also important to note that Ksp and solubility are two different things. Ksp 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 Ksp from solubility and vice versa.
For example: Cu2S3 (s) 3S2-(aq) + 2Cu3+(aq). The solubility for Cu2S3 is 1.0 X 10-5M. Calculate the Ksp.
Ksp = [S2-]3[Cu3+]2 because Ksp doesn’t include solids, you omit Cu2S3 from the expression. Because the coefficients are 3 and 2, the exponents are 3 and 2.
Because the mole ratio for Cu2S3 and S2- is 1-3 the solubility for S2-is 1.0 X 10-5 M Cu2S3 X 3mol S2-/ 1 mol Cu2S3 = 3.0 X 10-5M S2-. Because the mole ratio for Cu2S3 is 1-2 the solubility for Cu3+ is 1.0 X 10-5 M Cu2S3 X 2mol Cu3+/ 1 mol Cu2S3 = 2.0 X 10-5M Cu3+.
next you plug the values into the Ksp expression.
Ksp=[3.0 X 10-5M]3[2.0 X 10-5M]2= 1.08 X 10-3 units are not needed.
One can also calculate solubility from Ksp
For example: Sr(OH)2(s)—> 2OH–(aq) + Sr2+(aq). If the Ksp is 4.0 X 10-6 what is the solubility of OH–
First you write the expression. Ksp = [OH–]2[Sr2+]
then you make an I.C.E. table.
Sr(OH)2(s)—> 2OH–(aq) + Sr2+(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 10-8M. because there is a 2 in front of OH– we multiply it by 2. Thus the solubility of OH- is 2.0 X 10-8M.