Answer:
Ksp = 2.4 * 10^-13
Explanation:
Step 1: Data given
Molarity of NaIO3 = 0.10 M
The molar solubility of Pb(IO3)2 = 2.4 * 10^-11 mol/L
Step 2: The initial concentration
NaIO3 = 0.1M
Na+ = 0 M
2IO3- = 0 M
Step 3: The concentration at the equilibrium
All of the NaIO3 will react (0.1M)
At the equilibrium the concentration of NaIO3 = 0 M
The mol ratio is 1:1:1
The concentration of Na+ and IO3- is 0.1 M
Pb(IO3)2 → Pb^2+ + 2IO3^-
The concentration of Pb(IO3)2 can be written as X
The concentration of Pb^2+ can be written as X
The concentration of 2IO3^- can be written as 2X
Ksp = (Pb^2+)(IO3^-)²
⇒ with (Pb^2+) = 2.4*10^-11
⇒ with (IO3^-) = 2x from the Pb(IO3)2 and 0.1M from the NaIO3.
⇒The total (IO3^-) = 2x + 0.1 and we assume that x is so small that we can neglect it.
Ksp = (2.4 *10^-11)*(0.1)²
Ksp = 2.4 * 10^-13
