What mass of Ag2CO3 would be found in 1.4 L of a saturated solution if the Ksp of Ag2CO3 is 8.2 x 10-12?
Write the ionization equation (solubility) equation for Ag2CO3.
Let y = solubility of Ag2CO3.
Follow with concns of Ag^+ and CO3^=
Substitute into Ksp expression and solve for y. That will be mols/L. multiply by molar mass to find grams/L and multiply that by 1.4 to determine solubility in 1.4 L. Post your work if you get stuck.
To calculate the mass of Ag2CO3 in a saturated solution, we need to use the equilibrium expression for the solubility product constant (Ksp) and the formula mass of Ag2CO3.
The equilibrium expression for Ag2CO3 is:
Ag2CO3 ⇌ 2Ag+ + CO3^2-
The solubility product constant (Ksp) expression is:
Ksp = [Ag+]^2 * [CO3^2-]
We are given that the Ksp of Ag2CO3 is 8.2 x 10^-12.
Let's assume that 'x' represents the molar solubility of Ag2CO3 in moles per liter.
Then the concentration of Ag+ ions and CO3^2- ions in the solution is also 'x' (since the stoichiometric coefficient of Ag2CO3 is 1:1:1).
Using the Ksp expression, we can write:
Ksp = (x)^2 * (x)
8.2 x 10^-12 = x^3
To solve for 'x', we can take the cube root of both sides:
x = (8.2 x 10^-12)^(1/3)
Now we have the molar solubility of Ag2CO3, which is the same as the concentration of Ag+ ions and CO3^2- ions in the solution.
To find the mass of Ag2CO3, we can use the formula:
mass = molar solubility * molar mass
The molar mass of Ag2CO3 is:
2(Ag) + 1(C) + 3(O) = 2(107.87 g/mol) + 12.01 g/mol + 3(16.00 g/mol) = 275.87 g/mol
Now we can calculate the mass of Ag2CO3:
mass = x (molar solubility) * molar mass
mass = (8.2 x 10^-12)^(1/3) * 275.87 g/mol
By plugging in the given values, we can calculate the mass of Ag2CO3 in the 1.4 L saturated solution.