At 25 °C only 0.0580 mol of the generic salt AB3 is soluble in 1.00 L of water. What is the Ksp of the salt at 25 °C?
........AB3 ==> A^3+ + 3B^-
.........x.......x.......3x
x = 0.0580 = (A^3+)
3x = 3*0.0580 = (B^-)
Ksp = (A^3+)(B^-)^3
Ksp = (0.0580)(3*0.0580)^3 = ?
To find the Ksp of the salt AB3 at 25 °C, we need to use the information given about its solubility.
First, let's write the balanced chemical equation for the dissolution of AB3:
AB3(s) ⟷ A+(aq) + 3B-(aq)
The solubility of AB3 is given as 0.0580 mol/L. This means that for every 1 L of water, 0.0580 moles of AB3 dissolve. Since AB3 dissociates into one A+ ion and three B- ions, the equilibrium expression can be written as follows:
Ksp = [A+] * [B-]^3
Since the concentration of A+ is 0.0580 mol/L and the concentration of B- is 3 times that, we can substitute these values:
Ksp = (0.0580 mol/L) * (3 * 0.0580 mol/L)^3
Simplifying this equation:
Ksp = (0.0580 mol/L) * (3^3) * (0.0580 mol/L)^3
Ksp = 0.0580 * 3^3 * 0.0580^3 mol^4/L^4
Ksp ≈ 1.44 × 10^-6 mol^4/L^4
Therefore, the Ksp of the salt AB3 at 25 °C is approximately 1.44 × 10^-6 mol^4/L^4.
To find the solubility product constant (Ksp) of the salt AB3, we need to use the given information. Ksp is a measure of the solubility of a compound in water and is constant at a given temperature.
In this case, we know that at 25 °C, only 0.0580 moles of the salt AB3 is soluble in 1.00 L of water. We can use this solubility value to calculate the Ksp of the salt.
The equation for the dissolution of the salt AB3 is:
AB3 (s) ⇌ A3+ (aq) + 3B- (aq)
The molar solubility of AB3 can be defined as x mol/L. From the balanced equation, we can see that for every 1 mol of AB3 that dissolves, we get 1 mol of A3+ ions and 3 mol of B- ions. Therefore, the concentration of A3+ ions and B- ions in the solution is 1x and 3x, respectively.
Now, we can write the expression for the solubility product constant, Ksp, using the concentrations of the ions:
Ksp = [A3+]^1 * [B-]^3
Substituting the concentrations in terms of x:
Ksp = (1x)^1 * (3x)^3
= 27x^4
Given that the solubility of AB3 is 0.0580 mol/L, we have:
0.0580 mol/L = x
Substituting this value back into the Ksp expression:
Ksp = 27(0.0580)^4
Calculating this expression, we get:
Ksp ≈ 9.80 x 10^-12
Therefore, the Ksp of the salt AB3 at 25 °C is approximately 9.80 x 10^-12.