Calculate the Ksp for silver sulfate if the solubility of Ag2SO4 in pure water is 4.5 g/L.
Ag2SO4(s) <--> 2Ag^+(aq) + SO4^2-(aq)
mols Ag2SO4 = grams/molar mass - ? and that is the M since it is in 1 L. I estimate that to be 0.015 M
........Ag2SO4 <--> 2Ag^+ + SO4^2
I.......solid........0.......0
C.......solid........2x......x
E.......solid........2x......x
Substitute the E line into the Ksp expression and solve for Ksp.
To calculate the Ksp (solubility product constant) for silver sulfate (Ag2SO4), we need to use the solubility information provided.
The solubility of Ag2SO4 in pure water is given as 4.5 g/L. This means that at equilibrium, 4.5 grams of Ag2SO4 dissolves in one liter (L) of water.
Using the balanced chemical equation for the dissociation of Ag2SO4:
Ag2SO4(s) <--> 2Ag+(aq) + SO4^2-(aq)
We can see that for every 1 mole of Ag2SO4 that dissolves, it produces 2 moles of Ag+ ions and 1 mole of SO4^2- ions.
The molar mass of Ag2SO4 is:
Ag2SO4 = (2 x Ag) + (1 x S) + (4 x O)
= (2 x 107.87 g/mol) + (1 x 32.06 g/mol) + (4 x 16.00 g/mol)
= 311.80 g/mol
To calculate the concentration (in mol/L) of Ag+ ions (since there are 2 moles of Ag+ ions produced for every 1 mole of Ag2SO4 dissolved):
Concentration of Ag+ ions = (2 moles Ag+ ions / 311.80 g Ag2SO4) x (4.5 g Ag2SO4 / 1 L)
= (2/311.80) x (4.5/1) mol/L
= 0.02897 mol/L
Similarly, the concentration of SO4^2- ions is equal to:
Concentration of SO4^2- ions = (1 mole SO4^2- ions / 311.80 g Ag2SO4) x (4.5 g Ag2SO4 / 1 L)
= (1/311.80) x (4.5/1) mol/L
= 0.01449 mol/L
Therefore, at equilibrium, the concentrations of Ag+ and SO4^2- ions are 0.02897 mol/L and 0.01449 mol/L, respectively.
The solubility product constant (Ksp) for Ag2SO4 is the product of the concentrations of its dissociated ions at equilibrium:
Ksp = [Ag+]^2 * [SO4^2-]
= (0.02897 mol/L)^2 * (0.01449 mol/L)
= 0.00002459 mol^3/L^3
Therefore, the Ksp for silver sulfate (Ag2SO4) is approximately 0.00002459 mol^3/L^3.
To calculate the Ksp (solubility product constant) for silver sulfate (Ag2SO4), you need the balanced chemical equation and the molar solubility.
The balanced chemical equation is:
Ag2SO4(s) ⇌ 2Ag^+(aq) + SO4^2-(aq)
The molar solubility of Ag2SO4 is given as 4.5 g/L. We can convert this to moles per liter (mol/L) using the molar mass of Ag2SO4.
The molar mass of Ag2SO4 is:
Ag = 107.87 g/mol
S = 32.07 g/mol
O = 16.00 g/mol
Molar mass of Ag2SO4 = (2 * Ag) + S + (4 * O) = (2 * 107.87) + 32.07 + (4 * 16.00) = 311.82 g/mol
To calculate the molar solubility, we divide the given solubility by the molar mass:
Molar solubility = 4.5 g/L / 311.82 g/mol ≈ 0.0145 mol/L
The molar solubility of Ag2SO4 is 0.0145 mol/L.
Now, we can use the balanced chemical equation to establish the stoichiometry of the dissolved ions. According to the equation, 1 molecule of Ag2SO4 produces 2 Ag^+ ions and 1 SO4^2- ion.
So at equilibrium, the concentration of Ag^+ ions would be 2 * (0.0145 mol/L) = 0.029 mol/L.
The concentration of SO4^2- ions would be 1 * (0.0145 mol/L) = 0.0145 mol/L.
Finally, we can calculate the Ksp using the concentrations of the ions:
Ksp = [Ag^+]^2 * [SO4^2-] = (0.029 mol/L)^2 * (0.0145 mol/L) = 1.261 x 10^-5.
Therefore, the Ksp for silver sulfate (Ag2SO4) is approximately 1.261 x 10^-5.