A 40.0-mL solution contains 0.035 M barium chloride (BaCl2). What is the minimum concentration of sodium sulfate (Na2SO4) required in the solution to produce a barium sulfate (BaSO4) precipitate? The solubility product for barium sulfate is
Ksp = 1.1 ✕ 10−10.
BaSO4 ==> Ba^2+ + [SO4]^2-
Ksp = (Ba^2+) [SO4]^2-
You know Kap and you know (Ba^2+) from the problem. Solve for sulfate ion.concentration in mols/L = M. In this case th 40.0 mL doesn't matter.
To determine the minimum concentration of sodium sulfate (Na2SO4) required to produce a barium sulfate (BaSO4) precipitate, we need to use the solubility product constant (Ksp) for barium sulfate.
The balanced chemical equation for the reaction between barium chloride (BaCl2) and sodium sulfate (Na2SO4) is:
BaCl2 + Na2SO4 -> BaSO4 + 2NaCl
From the balanced equation, we can see that the molar ratio between BaCl2 and BaSO4 is 1:1. This means that every 1 mole of BaCl2 will produce 1 mole of BaSO4.
Given that the initial volume of the solution is 40.0 mL, and the concentration of BaCl2 is 0.035 M, we can calculate the number of moles of BaCl2:
moles of BaCl2 = volume x concentration
= 0.040 L x 0.035 mol/L
= 0.0014 mol
Since the molar ratio between BaCl2 and BaSO4 is 1:1, the number of moles of BaSO4 produced will also be 0.0014 mol.
The solubility product constant (Ksp) for BaSO4 is given as 1.1 x 10^(-10). This represents the product of the concentration of the barium ions (Ba^2+) and sulfate ions (SO4^2-) in the solution when BaSO4 is at the point of saturation.
Using stoichiometry, we can determine the concentration of Ba^2+ in the solution, which is equivalent to the concentration of BaSO4:
concentration of Ba^2+ = moles of BaSO4 / volume
= 0.0014 mol / 0.040 L
= 0.035 M
Since the concentration of Ba^2+ is the same as the concentration of BaSO4, we can substitute this value into the solubility product constant equation and solve for the minimum concentration of Na2SO4:
Ksp = [Ba^2+] x [SO4^2-]
1.1 x 10^(-10) = (0.035 M) x ([SO4^2-])
[SO4^2-] = (1.1 x 10^(-10)) / (0.035 M)
[SO4^2-] = 3.14 x 10^(-9) M
Therefore, the minimum concentration of sodium sulfate (Na2SO4) required in the solution to produce a barium sulfate (BaSO4) precipitate is approximately 3.14 x 10^(-9) M.
To determine the minimum concentration of sodium sulfate (Na2SO4) required in the solution to produce a barium sulfate (BaSO4) precipitate, we need to consider the stoichiometry of the reaction and the solubility product.
The balanced chemical equation for the reaction between barium chloride (BaCl2) and sodium sulfate (Na2SO4) is:
BaCl2 + Na2SO4 -> BaSO4 + 2NaCl
From the equation, we can see that 1 mole of barium chloride reacts with 1 mole of sodium sulfate to produce 1 mole of barium sulfate.
Given that the initial concentration of barium chloride (BaCl2) is 0.035 M, the number of moles of barium chloride can be calculated by multiplying the concentration by the volume:
moles of BaCl2 = concentration of BaCl2 * volume = 0.035 M * 0.040 L = 0.0014 moles
Since the stoichiometry is 1:1 between barium chloride and barium sulfate, the maximum number of moles of barium sulfate that can be precipitated is also 0.0014 moles.
Now, we can use the solubility product (Ksp) for barium sulfate (BaSO4) to determine the minimum concentration of sodium sulfate required. The solubility product expression for barium sulfate is:
Ksp = [Ba2+][SO42-]
Since the stoichiometry of the reaction is 1:1, the concentration of barium sulfate is equal to the concentration of barium ions:
[Ba2+] = concentration of BaSO4
Substituting the values into the solubility product expression:
Ksp = [Ba2+][SO42-]
1.1 x 10^-10 = (concentration of BaSO4)(concentration of SO42-)
We know that the concentration of BaSO4 (which is the same as the concentration of Ba2+) is 0.0014 moles divided by the total volume of the solution (40.0 mL or 0.040 L):
0.0014 moles / 0.040 L = 0.035 M
Substituting the values into the solubility product expression:
1.1 x 10^-10 = (0.035 M)(concentration of SO42-)
Rearranging the equation to solve for the concentration of sulfate ions (SO42-):
concentration of SO42- = (1.1 x 10^-10) / (0.035 M)
concentration of SO42- = 3.14 x 10^-9 M
Therefore, the minimum concentration of sodium sulfate required in the solution to produce a barium sulfate precipitate is 3.14 x 10^-9 M.