For a binary salt AB the Ksp is 4 × 10–10. The molar solubility of AB is :-2 × 10–5 M5 × 10–5 M2 × 10–10 M4 × 10–5 M
Question
For a binary salt AB the Ksp is 4 × 10–10. The molar solubility of AB is :-
- 2 × 10–5 M
- 5 × 10–5 M
- 2 × 10–10 M
- 4 × 10–5 M
Solution
The solubility product constant, Ksp, is used to describe the equilibrium between a solid and its constituent ions in a solution. The ions are produced when the solid dissolves.
For a binary salt AB, it dissociates into its ions as follows:
AB(s) ↔ A+(aq) + B-(aq)
The equilibrium expression for this reaction is:
Ksp = [A+][B-]
Since the salt dissociates into equal amounts of A+ and B- ions, we can say that [A+] = [B-] = x, where x is the molar solubility of the salt.
Substituting these into the equilibrium expression gives:
Ksp = x^2
Given that Ksp = 4 × 10–10, we can solve for x (the molar solubility) as follows:
x^2 = 4 × 10–10 x = sqrt(4 × 10–10) x = 2 × 10–5 M
So, the molar solubility of AB is 2 × 10–5 M.
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