At 25 °C only 0.0390 mol of the generic salt AB3 is soluble in 1.00 L of water. What is the Ksp of the salt at 25 °C?
To find the value of Ksp for the salt AB3 at 25 °C, we need to use the given solubility information. The solubility of AB3 is given as 0.0390 mol/L, which means that 0.0390 moles of AB3 dissolve in 1.00 L of water.
The generic formula AB3 suggests that, for every mole of AB3 that dissolves, it produces one mole of A ions (A+) and three moles of B ions (B3-). Therefore, the solubility of AB3 can be expressed as follows:
Solubility (mol/L) = [A+] × [B3-]^3
Given that the solubility of AB3 is 0.0390 mol/L, we can establish the following relationship:
0.0390 = [A+] × [B3-]^3
Since the concentration of A ions is equal to the concentration of B ions (assuming 1:1 stoichiometry), we can rewrite the equation as follows:
0.0390 = [A+] × ([A+]^3)^3
0.0390 = [A+]^4
To solve for [A+], we need to take the fourth root of both sides of the equation:
[A+] = (0.0390)^(1/4)
[A+] = 0.5489 mol/L
Now that we have determined the concentration of A ions, we can find the concentration of B ions. Since the ratio is 1:3, the concentration of B ions ([B3-]) is three times the concentration of A ions:
[B3-] = 3 × [A+]
[B3-] = 3 × 0.5489
[B3-] = 1.6468 mol/L
Finally, to calculate the value of Ksp, we multiply the concentrations of the ions:
Ksp = [A+] × [B3-]³
Ksp = 0.5489 × (1.6468)^3
Ksp = 5.98
Therefore, the value of Ksp for the salt AB3 at 25 °C is approximately 5.98.
........AB3 ==>A + 3B^3-
I......solid...0....0
C....solid...0.039..3*0.039
E....solid...0.039..0.117
Ksp = (A)(B^3-)^3
Substitute the E line into Ksp and solve.