What is the value of Ksp for Ag2SO4
if 5.70 g is soluble in 1.00 L of water?
To find the value of Ksp (solubility product constant) for Ag2SO4, we need the equation for the dissociation of Ag2SO4 in water. The equation is:
Ag2SO4(s) ⇌ 2Ag+(aq) + SO4^2-(aq)
The Ksp expression for this equilibrium is:
Ksp = [Ag+]^2 * [SO4^2-]
Given that 5.70 g of Ag2SO4 is soluble in 1.00 L of water, we need to convert this mass to moles.
First, calculate the molar mass of Ag2SO4:
Ag: 2 * Atomic mass of Ag = 2 * 107.87 g/mol = 215.74 g/mol
S: Atomic mass of S = 32.06 g/mol
O: 4 * Atomic mass of O = 4 * 16.00 g/mol = 64.00 g/mol
Total molar mass of Ag2SO4 = 215.74 g/mol + 32.06 g/mol + 64.00 g/mol = 311.8 g/mol
Next, calculate the number of moles of Ag2SO4:
moles = mass / molar mass
moles = 5.70 g / 311.8 g/mol = 0.0183 mol
Since Ag2SO4 dissociates into 2 Ag+ ions and 1 SO42- ion, the concentration of Ag+ ions and SO42- ions in the saturated solution will be twice the concentration of Ag2SO4:
[Ag+] = 2 * 0.0183 mol/L = 0.0366 mol/L
[SO4^2-] = 0.0183 mol/L
Finally, substitute the concentrations into the Ksp expression to find the Ksp value:
Ksp = [Ag+]^2 * [SO4^2-]
Ksp = (0.0366 mol/L)^2 * (0.0183 mol/L)
Ksp = 1.97 x 10^-5 mol^3/L^3
Therefore, the value of Ksp for Ag2SO4 is 1.97 x 10^-5 mol^3/L^3.