Answer:
a. Ksp = 4s³
b. 5.53 Ć 10ā“ mol³/dmā¹
Explanation:
a. Obtain an expression for the solubility product of AB2(S),in terms of s.
ABā dissociates to give
ABā ā A²⺠+ 2Bā»
Since 1 mole of ABā gives 1 mole of A and 2 moles of B, we have the mole ratio as
ABā ā A²⺠+ 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 A²⺠+ 2Bā»
[s] Ā Ā Ā Ā [s] Ā Ā [2s]
So, the solubility product Ksp = [A²āŗ][Bā»]²
= (s)(2s)²
= s(4s²)
= 4s³
b. Calculate the Ksp of ABā, given that solubility is 2.4 Ć 10³ mol/dm³
Given that the solubility of AB is 2.4 Ć 10³ mol/dm³ and the solubility product Ksp = [A²āŗ][Bā»]² = 4s³ where s = solubility of AB = 2.4 Ć 10³ mol/dm³
Substituting the value of s into the equation, we have
Ksp = 4s³
= 4(2.4 à 10³ mol/dm³)³
= 4(13.824 Ć 10³ mol³/dmā¹)
= 55.296 Ć 10³ mol³/dmā¹
= 5.5296 Ć 10ā“ mol³/dmā¹
ā 5.53 Ć 10ā“ mol³/dmā¹
Ksp = 5.53 Ć 10ā“ mol³/dmā¹