Correct task : At 25°C, only 0.0140 mol of the generic salt AB3 is soluble in 1.00 L of water. What is the Ksp of this salt at 25°C.
Solution:
When the salt dissolves, it dissociates as follows:
AB3 --> A3+ + 3B⁻
--S--------S-------3S--
The molar solubility (S) is the number of moles that can dissolve in 1 L of solution.
Molar solubility of salt is (0.0140 mol) / (1.00 L) = 0.0140 mol/L.
According to the dissociation equation:
Solubility of A3+ is 0.0140 mol/L and solubility of B⁻ is 3×0.0140 mol= 0.0420 mol/L.
[A3+] = 0.0140 mol/L.
[B⁻] = 0.0420 mol/L.
Ksp is the solubility product constant and calculated as follows:
Ksp(AB3) = [A3+] × [B⁻]3
Ksp(AB3) = [0.0140] × [0.0420]3
Ksp(AB3) = 10.37×10-7.
Ksp of this salt is 1.04×10-6.
Answer: 1.04×10-6 is the Ksp of this salt at 25°C.