At 25 °C only 0.0570 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 the salt AB at 25 °C, we need to use the given information that 0.0570 mol of AB dissolves in 1.00 L of water.
The general equation for the solubility equilibrium of AB is given by:
AB ⇌ A+ + B-
We can assume that AB dissociates completely into A+ and B- ions.
The solubility of AB can be represented by the equation:
Ksp = [A+][B-]
Since the concentration of A+ and B- ions in solution will be equal, we can represent them as [A+]^2.
Given that the concentration of AB is 0.0570 mol/L, we can substitute this value into our equation:
Ksp = (0.0570 mol/L)^2
Ksp = 0.003249 mol^2/L^2
Therefore, the Ksp of the salt AB at 25 °C is 0.003249 mol^2/L^2.
To find the Ksp (solubility product constant) of the salt AB at 25 °C, we need to use the given information about its solubility.
The solubility of a substance is defined as the amount of the substance that dissolves in a given amount of solvent to form a saturated solution. In this case, the solubility of AB is given as 0.0570 mol/L.
The Ksp expression for the salt AB can be written as follows:
AB ⇌ A+ + B-
Where AB represents the salt, A+ represents the cation, and B- represents the anion.
The Ksp expression is defined as the product of the concentrations of the cation and anion raised to the power of their respective stoichiometric coefficients. Since the stoichiometry of the dissociation is 1:1, the Ksp expression simplifies to:
Ksp = [A+][B-]
Now, we need to determine the concentrations of A+ and B- ions in the saturated solution. Since AB is soluble, it dissociates completely into its ions:
[A+] = [B-] = solubility of AB = 0.0570 mol/L
Substituting the values into the Ksp expression, we get:
Ksp = (0.0570 mol/L)(0.0570 mol/L) = 0.003249
Therefore, the Ksp of the salt AB at 25 °C is 0.003249.
solubility = 0.0570 mol/L = 0.0570M
........AB ==> A^+ + B^-
I......solid.. 0.....0
C......-x......x.....x
E......solid...x.....x
Ksp = (A^+)(B^-)
Ksp = (x)(x)
Substitute the value of x and calculate Ksp.
All we know about salt AB is that the oxidation states of A and B are the same; therefore, it could be
A^+ and B^- or
A^2+ and B^2- or-
A^3+ and B^3-
but whatever it is Ksp is done the same way.