Final answer:
The molar solubility of lanthanum iodate, La(IO3)3, is calculated by setting up the Ksp expression and substituting equilibrium concentrations based on the solubility. After solving for s, the molar solubility is found to be 1.40×10−3 M.
Step-by-step explanation:
To calculate the molar solubility s of lanthanum iodate, La(IO3)3, given its solubility product constant (Ksp) of 7.50×10−12, we must consider the dissolution reaction:
La(IO3)3 (s) → La3+ (aq) + 3 IO3− (aq)
If the solubility is s, the equilibrium concentrations of the ions will be:
[La3+] = s
[IO3−] = 3s
Using the expression for Ksp:
Ksp = [La3+][IO3−]3
We can substitute the concentrations based on solubility s into the Ksp expression:
Ksp = s(3s)3 = 27s4
Plugging in the Ksp value given:
7.50×10−12 = 27s4
To solve for s, we take the fourth root:
s = −(rac{7.50×10−12}{27})¼
s = 7.50 < 1027×10−12
After calculating this value, we find:
s = 1.40×10−3 M
This is the molar solubility of La(IO3)3.