Answer:
a. Ksp = 4sĀ³
b. 5.53 Ć 10ā“ molĀ³/dmā¹
Explanation:
a. Obtain an expression for the solubility product of AB2(S),in terms of s.
ABā dissociates to give
ABā ā AĀ²āŗ + 2Bā»
Since 1 mole of ABā gives 1 mole of A and 2 moles of B, we have the mole ratio as
ABā ā AĀ²āŗ + 2Bā»
1 : 1 : 2
Since the solubility of ABā is s, then the solubility of A is s and that of B is 2s
So, we have
ABā ā AĀ²āŗ + 2Bā»
[s] Ā Ā Ā Ā [s] Ā Ā [2s]
So, the solubility product Ksp = [AĀ²āŗ][Bā»]Ā²
= (s)(2s)Ā²
= s(4sĀ²)
= 4sĀ³
b. Calculate the Ksp of ABā, given that solubility is 2.4 Ć 10Ā³ mol/dmĀ³
Given that the solubility of AB is 2.4 Ć 10Ā³ mol/dmĀ³ and the solubility product Ksp = [AĀ²āŗ][Bā»]Ā² = 4sĀ³ where s = solubility of AB = 2.4 Ć 10Ā³ mol/dmĀ³
Substituting the value of s into the equation, we have
Ksp = 4sĀ³
= 4(2.4 Ć 10Ā³ mol/dmĀ³)Ā³
= 4(13.824 Ć 10Ā³ molĀ³/dmā¹)
= 55.296 Ć 10Ā³ molĀ³/dmā¹
= 5.5296 Ć 10ā“ molĀ³/dmā¹
ā
5.53 Ć 10ā“ molĀ³/dmā¹
Ksp = 5.53 Ć 10ā“ molĀ³/dmā¹
