At 25 °C only 0.0990 mol of the generic salt AB is soluble in 1.00 L of water. What is the Ksp of the salt at 25 °C?
To find the Ksp (solubility product constant) of a salt, you need to know the equilibrium concentrations of its ions in a saturated solution.
The formula for the solubility product constant (Ksp) is:
Ksp = [A+]^a [B-]^b
Where [A+] is the concentration of the cation A+ in moles per liter (mol/L), [B-] is the concentration of the anion B- in mol/L, and a and b are the coefficients of the ions in the balanced equation for the dissolution of the salt.
In this case, the generic salt AB will dissociate into the cation A+ and the anion B-. However, since the equation for the dissolution of AB is not provided, we will assume it goes according to:
AB (s) ↔ A+ (aq) + B- (aq)
From the given information, we know that at 25 °C, only 0.0990 mol of AB is soluble in 1.00 L of water. This means the equilibrium concentration of AB is 0.0990 mol/L.
However, since AB is a sparingly soluble salt, we assume that it will completely dissociate in water, so we can consider the equilibrium concentrations of A+ and B- to be equal to each other and equal to x mol/L.
Therefore, the equation becomes:
AB (s) ↔ x (aq) + x (aq)
Since AB is a 1:1 salt, both A+ and B- have coefficients of 1 in the dissolution equation.
Substituting the values into the Ksp expression:
Ksp = [A+]^1 [B-]^1 = x * x = x^2
Now, we need to solve for x. This can be done by using the solubility of AB, which is 0.0990 mol/L. Therefore, x = 0.0990 mol/L.
Now, substitute this value of x into the Ksp expression:
Ksp = (0.0990 mol/L)^2 = 0.009801 mol^2/L^2
Therefore, the Ksp of the generic salt AB at 25 °C is 0.009801 mol^2/L^2.
To find the Ksp (solubility product constant) of the salt AB, we need the concentration of the dissolved ions.
Since the salt AB dissociates into A+ and B- ions, we can assume that the concentration of A+ ions and B- ions is the same.
Given that 0.0990 mol of the salt AB is dissolved in 1.00 L of water, we can calculate the concentration of A+ and B- ions using the molar concentration formula:
Concentration (C) = Number of moles (n) / Volume (V)
Concentration of A+ and B- ions = 0.0990 mol / 1.00 L = 0.0990 M
Since the concentration of A+ ions and B- ions is the same, we can represent it as [A+] and [B-].
Now, we can write the Ksp expression for the salt AB:
Ksp = [A+][B-]
Substituting the concentration values we calculated:
Ksp = (0.0990 M) x (0.0990 M)
Ksp = 0.009801 M²
Therefore, the Ksp of the salt AB at 25 °C is 0.009801 M².
.........AB==> A^+ + B^-
I.......solid..0......0
C.......solid..x......x
E.......solid..x......x
Ksp = (A^+)(B^-) = (x)(x)
The problem tell you 0.0990 mol soluble in 1L; therefore M = x = 0.0990/1
Substitute and solve for Ksp.
Note: the salt could be
AB ==> A^2+ + B^2- or
AB ==> A^3+ + B^3-
but none of that will change Ksp.