If 1 liter of the Pentland field formation water (see lecture power-point) is mixed with equal volume of seawater and noting that BaSO4 solubility under that condition is 2 mg/liter, how much BaSO4 precipitate would be there? (Hint: watch out for which ion is the limiting factor)( NOTE: Ksp (BaSO4) is 1x10-10 Atomic weight of Ba is 137, and molecular weight of SO4 is 96;

To determine how much BaSO4 precipitate would form when 1 liter of the Pentland field formation water is mixed with an equal volume of seawater, we need to consider the solubility of BaSO4 and the limiting factor in the reaction.

The solubility of BaSO4 is given as 2 mg/liter, which means that 2 mg of BaSO4 will dissolve in 1 liter of water under the specified conditions. However, we need to determine if there is enough Ba2+ or SO42- ions to form this amount of BaSO4.

The solubility product constant (Ksp) for BaSO4 is given as 1x10^-10. This means that the product of the concentrations of Ba2+ and SO42- ions should be equal to or less than this value for BaSO4 to remain in solution.

Let's start by calculating the concentration of Ba2+ and SO42- ions in the mixed solution.

The molecular weight of Ba is 137, so the concentration of Ba2+ ions in 1 liter of the mixed solution would be equal to the number of moles of Ba divided by the total volume in liters. Since we mixed 1 liter of Pentland field formation water with 1 liter of seawater, the total volume is 2 liters.

Concentration of Ba2+ = (Mass of Ba in grams / Molecular weight of Ba) / Total volume
Concentration of Ba2+ = (1 liter * 137 g/mol) / 2 liters
Concentration of Ba2+ = 68.5 g

The molecular weight of SO4 is 96, so the concentration of SO42- ions in 1 liter of the mixed solution would be equal to the number of moles of SO4 divided by the total volume in liters.

Concentration of SO42- = (Mass of SO4 in grams / Molecular weight of SO4) / Total volume
Concentration of SO42- = (2 mg * 1 gram / 1000 mg / 96 g/mol) / 2 liters
Concentration of SO42- = 0.0104 g

Now, let's calculate the product of the concentrations of Ba2+ and SO42- ions.

Product of concentrations = Concentration of Ba2+ * Concentration of SO42-
Product of concentrations = 68.5 g * 0.0104 g
Product of concentrations = 0.7112 g^2

Since the product of concentrations (0.7112 g^2) is much smaller than the solubility product constant (1x10^-10), there is no excess of either Ba2+ or SO42- ions in the mixed solution, and we can assume that both ions will fully react to form BaSO4.

Therefore, 2 mg of BaSO4 will precipitate in the mixed solution, as the reaction goes to completion without any limiting factors.