Write down the Ksp expression for Ag2CrO4. If the solubility of Ag2CrO4 is 0.0216 g/L, calculate the value of Ksp
See the CaF2 problem. Same principle.
The Ksp expression for a salt represents the solubility product constant, which is a measure of the extent to which a compound dissolves in a solution. For Ag2CrO4, the Ksp expression is:
Ksp = [Ag+]^2 * [CrO4^2-]
To calculate the value of Ksp, we need to determine the concentration of Ag+ and CrO4^2- ions in the solution. Given that the solubility of Ag2CrO4 is 0.0216 g/L, we can convert this mass into molar concentration.
First, we need to find the molar mass of Ag2CrO4:
Ag: atomic mass = 107.87 g/mol
Cr: atomic mass = 52.00 g/mol
O: atomic mass = 16.00 g/mol
Molar mass of Ag2CrO4:
(2 * 107.87) + 52.00 + (4 * 16.00) = 331.73 g/mol
To convert the solubility from grams per liter to moles per liter, we divide the mass by the molar mass:
0.0216 g/L / 331.73 g/mol = 6.51 × 10^-5 mol/L
Since Ag2CrO4 dissociates into two Ag+ ions and one CrO4^2- ions, the concentrations can be determined by multiplying the solubility by the dissociation coefficients:
[Ag+] = 2 * 6.51 × 10^-5 mol/L = 1.30 × 10^-4 mol/L
[CrO4^2-] = 1 * 6.51 × 10^-5 mol/L = 6.51 × 10^-5 mol/L
Now we can substitute these values into the Ksp expression:
Ksp = (1.30 × 10^-4)^2 * (6.51 × 10^-5) = 1.17 × 10^-11
Therefore, the value of Ksp for Ag2CrO4 is 1.17 × 10^-11.
To write down the solubility product constant (Ksp) expression for Ag2CrO4, we need to balance and express the chemical equation for the dissociation of Ag2CrO4 in water.
The balanced equation for the dissociation of Ag2CrO4 can be written as:
Ag2CrO4 (s) ⇌ 2Ag+ (aq) + CrO4^2- (aq)
The Ksp expression can be written by multiplying the concentrations of the dissociated ions raised to their respective stoichiometric coefficients:
Ksp = [Ag+]^2 [CrO4^2-]
Now, to calculate the value of Ksp using the provided solubility, we need to convert the solubility from grams per liter to moles per liter.
Given:
Solubility of Ag2CrO4 = 0.0216 g/L
The molar mass of Ag2CrO4 can be calculated by adding the atomic masses:
Ag2CrO4 = (2 * atomic mass of Ag) + atomic mass of Cr + (4 * atomic mass of O)
Ag2CrO4 = (2 * 107.87 g/mol) + 52.00 g/mol + (4 * 16.00 g/mol)
Ag2CrO4 = 331.74 g/mol
To convert grams to moles, we can use the molar mass:
0.0216 g/L * (1 mol / 331.74 g) = 6.51 x 10^-5 mol/L
Since Ag2CrO4 dissociates into 2 Ag+ and 1 CrO4^2-, the concentration of Ag+ and CrO4^2- ions will be twice the initial concentration:
[Ag+] = 2 * (6.51 x 10^-5 mol/L) = 1.30 x 10^-4 mol/L
[CrO4^2-] = 1 * (6.51 x 10^-5 mol/L) = 6.51 x 10^-5 mol/L
Substituting these values into the Ksp expression:
Ksp = (1.30 x 10^-4 mol/L)^2 * (6.51 x 10^-5 mol/L)
Ksp = 1.55 x 10^-9
Therefore, the value of the solubility product constant (Ksp) for Ag2CrO4 is 1.55 x 10^-9.