A generic salt, AB, has a molar mass of 293 g/mol and a solubility of 2.20 g/L at 25 °C. What is the Ksp of this salt at 25 °C?
2.20g/293 = about 0.007 but you need to get a more accurate number.
.......AB ==> A + B
I....solid....0...0
C....solid....x...x
E....solid....x...x
The problem tells you that is is 0.0
07 mol/L. Substitute this into the Ksp expression and solve for Ksp.
To determine the solubility product constant (Ksp), we need to use the molar mass of the salt, the solubility, and the balanced chemical equation for the dissolution of the salt in water.
Given:
Molar mass of salt (AB) = 293 g/mol
Solubility = 2.20 g/L
To find the Ksp, we need to relate the concentration of the dissolved ions to the molar solubility of the salt. Assuming the balanced chemical equation for the dissolution of AB is:
AB (s) ⇌ A+ (aq) + B- (aq)
The solubility of AB (g/L) can be converted into the concentration of the dissolved ions (mol/L) using the molar mass:
Molar solubility of AB (mol/L) = (solubility of AB (g/L)) / (molar mass of AB (g/mol))
Molar solubility of AB = (2.20 g/L) / (293 g/mol) = 0.00751 mol/L
Now, the Ksp expression for the dissolution reaction can be written as:
Ksp = [A+]^x * [B-]^y
Since the stoichiometric coefficient for A+ and B- is 1 (obtained from the balanced equation), we can simplify the Ksp expression to:
Ksp = [A+]*[B-]
At equilibrium, the concentration of A+ and B- is equal to the molar solubility of AB:
Ksp = (molar solubility of AB)^2
Ksp = (0.00751 mol/L)^2 = 5.63 × 10^-5 mol^2/L^2
Therefore, the Ksp of the salt AB at 25 °C is 5.63 × 10^-5 mol^2/L^2.