What is the maximum amount of barium sulfate that will dissolve in 600mL of 0.02M Na2SO4? Barium sulfate ksp=1.1*10^-10
To determine the maximum amount of barium sulfate that will dissolve in 600 mL of 0.02 M Na2SO4, we need to consider the solubility product constant (Ksp), which tells us the equilibrium concentration of the ions in a saturated solution.
The balanced equation for the dissolution of barium sulfate (BaSO4) is:
BaSO4(s) ⇌ Ba2+(aq) + SO4^2-(aq)
The Ksp expression for barium sulfate is:
Ksp = [Ba2+][SO4^2-]
Given that Ksp = 1.1 × 10^-10 and the concentration of Na2SO4 is 0.02 M, we will assume that it doesn't significantly affect the solubility of barium sulfate as it is a common ion effect.
To find the maximum amount of barium sulfate that will dissolve, we first need to calculate the maximum concentrations of Ba2+ and SO4^2- in a saturated solution using the Ksp expression. Since barium sulfate has a 1:1 stoichiometry, the concentrations of Ba2+ and SO4^2- will be equal at equilibrium.
Let's assume the concentration of dissolved BaSO4 is x M. Therefore, at equilibrium, the concentration of Ba2+ and SO4^2- will both be x M.
Using the Ksp expression:
Ksp = x * x
1.1 × 10^-10 = x^2
To solve for x, we take the square root of both sides:
√(1.1 × 10^-10) = x
x ≈ 1.05 × 10^-5 M
Now that we know the concentration of barium sulfate in a saturated solution, we can determine the maximum amount of barium sulfate that will dissolve in 600 mL of solution.
The maximum amount is given by the equation:
Maximum amount = concentration × volume
Maximum amount = (1.05 × 10^-5 M) × 0.6 L
Maximum amount = 6.3 × 10^-6 moles
Therefore, the maximum amount of barium sulfate that will dissolve in 600 mL of 0.02 M Na2SO4 is approximately 6.3 × 10^-6 moles.