A generic salt, AB2, has a molar mass of 159 g/mol and a solubility of 3.30 g/L at 25 °C. What is the Ksp of this salt at 25 °C?
I've done one of each type. You try this one on your own. Post your work if you get stuck.
To find the Ksp (Solubility Product Constant) of a salt, we need to use the solubility data given for the salt.
The solubility of the salt is given as 3.30 g/L. From this information, we can calculate the concentration of the salt in moles per liter (mol/L).
First, let's calculate the number of moles of the salt in 3.30 g using its molar mass:
Number of moles of AB2 = mass / molar mass
Number of moles of AB2 = 3.30 g / 159 g/mol
Now, let's convert the number of moles to moles per liter (mol/L):
Concentration of AB2 = number of moles / volume
Concentration of AB2 = (3.30 g / 159 g/mol) / 1 L
Concentration of AB2 = (3.30 / 159) mol/L
Now that we have the concentration of the salt, we can set up the equilibrium expression for the solubility reaction of AB2:
AB2 (s) ⇌ A^2+ (aq) + 2B^- (aq)
The solubility product expression for this reaction is:
Ksp = [A^2+][B^-]^2
Since the stoichiometry of the reaction is 1:2 (1 A^2+ ion and 2 B^- ions are produced), we can set up the equilibrium expression using the concentration of AB2:
Ksp = [A^2+][B^-]^2 = (concentration of AB2) * (concentration of B^-)^2
Now substitute the concentration of AB2:
Ksp = (3.30 / 159) * (concentration of B^-)^2
However, we don't know the concentration of B^-. To find that, we need to make use of the charge balance in the salt AB2. The salt AB2 is neutral, which means the amount of positive charge (A^2+) must be equal to the amount of negative charge (B^-):
A^2+ = 2B^-
Since the concentration of A^2+ is the same as the concentration of B^-, we can substitute the concentration of B^- with (concentration of AB2):
Ksp = (3.30 / 159) * (3.30 / 159)^2
Now, we can calculate the Ksp by evaluating this expression:
Ksp = (3.30 / 159) * (3.30 / 159)^2 = 1.35 x 10^-6 (approximately)
Therefore, the Ksp of the salt AB2 at 25 °C is 1.35 x 10^-6.