A generic salt, AB, has a molar mass of 313 g/mol and a solubility of 4.30 g/L at 25 °C. What is the Ksp of this salt at 25 °C?
To find the Ksp (solubility product constant) of the salt, you need to know the concentrations of the ions in the solution formed by the salt when it dissolves.
The generic salt AB can be represented by the formula AB(s) ↔ A+(aq) + B-(aq), where A+ and B- are the ions formed from the salt.
Given that the molar mass of the salt (AB) is 313 g/mol, you can assume that 1 mole of AB produces 1 mole each of A+ and B- ions.
The solubility of the salt is given as 4.30 g/L at 25 °C. Since the molar mass is 313 g/mol, you can calculate the molarity (M) of the salt using the given solubility as follows:
Molarity (M) = (mass of solute in grams) / (molar mass in g/mol) = (4.30 g/L) / (313 g/mol)
Molarity (M) = 0.0137 mol/L
Now that you know the concentration of the ions in the solution, you can set up the expression for the solubility product constant (Ksp) based on the balanced equation:
Ksp = [A+][B-]
Since the concentration of both A+ and B- are equal, let's assume x mol/L for each ion:
Ksp = (x)(x) = x^2
Substituting the value of the concentration (0.0137 mol/L) into the equation:
Ksp = (0.0137 mol/L)^2
Ksp = 1.88 x 10^-4 (mol/L)^2
Therefore, the Ksp of the salt (AB) at 25 °C is 1.88 x 10^-4 (mol/L)^2.
To determine the Ksp (solubility product constant) of a salt, you need to know the molar mass and solubility of the salt. The Ksp represents the equilibrium constant for the dissolution of the salt in water.
First, let's write the balanced equation for the dissociation of the salt in water:
AB(s) ⟶ A+(aq) + B-(aq)
The molar mass of AB is given as 313 g/mol, which means that one mole of AB weighs 313 grams.
The solubility of AB at 25 °C is given as 4.30 g/L, which means that for every liter of water, you can dissolve 4.30 grams of AB.
Now, we can use these values to calculate the Ksp.
Ksp is determined by the concentration of the ions in solution at equilibrium. Since AB dissociates into one A+ ion and one B- ion, the concentration of each ion at equilibrium will be the same.
Let's assume the solubility of AB is 's' moles per liter.
The concentration of A+ and B- ions will be 's' moles per liter.
Therefore, the Ksp expression for AB will be:
Ksp = [A+][B-]
Since the concentration of A+ and B- is the same and equal to 's', we can rewrite the Ksp expression as:
Ksp = [A+]^2
To determine the concentration of A+ ions, we need to convert the solubility in grams per liter (g/L) to moles per liter (mol/L).
The molar mass of AB is 313 g/mol, and you can dissolve 4.30 g of AB in one liter of water. Therefore, the concentration (in moles per liter) of AB is:
s = (4.30 g/L) / (313 g/mol) = 0.0137 mol/L
Now, we can substitute this value back into the Ksp expression:
Ksp = [A+]^2 = (s)^2 = (0.0137 mol/L)^2 = 1.88 x 10^-4 (mol/L)^2
Therefore, the Ksp of the salt, AB, at 25 °C is 1.88 x 10^-4 (mol/L)^2.