Calcium Bromide Cooling And Phosphate Precipitation A Stoichiometry Challenge

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Embark on a fascinating journey into the realm of chemistry as we dissect a complex problem involving the cooling of a calcium bromide solution and the subsequent precipitation of calcium ions using sodium phosphate. This intricate scenario requires a meticulous step-by-step approach, leveraging our knowledge of solubility, stoichiometry, and chemical reactions. Let's unravel the layers of this problem and arrive at the solution.

Initial State: Saturated Calcium Bromide Solution

Our starting point is a 1000 g solution of 20% calcium bromide (CaBr2). This means that 20% of the solution's mass is attributed to CaBr2, while the remaining 80% is the solvent, presumably water. To kick things off, let's calculate the mass of CaBr2 present in the initial solution:

Mass of CaBr2 = 1000 g (solution) * 0.20 = 200 g

Consequently, the mass of water in the solution is:

Mass of H2O = 1000 g (solution) - 200 g (CaBr2) = 800 g

The density of the solution is given as 1.25 g/mL, which will be crucial later if we need to convert between mass and volume. For now, we'll stick with mass-based calculations.

The Cooling Process and Precipitation

Upon cooling, 116 g of CaBr2.5H2O precipitates out of the solution. This is a crystalline hydrate, meaning that each mole of CaBr2 is associated with five moles of water molecules. The formation of this hydrate significantly alters the composition of the remaining solution, as both CaBr2 and water are removed from it.

To accurately determine the impact of this precipitation, we need to calculate the moles of CaBr2.5H2O that precipitated:

Molar mass of CaBr2.5H2O = 40.08 (Ca) + 2 * 79.90 (Br) + 5 * (2 * 1.01 (H) + 16.00 (O)) = 295.97 g/mol

Moles of CaBr2.5H2O = 116 g / 295.97 g/mol ≈ 0.392 mol

Since each mole of CaBr2.5H2O contains one mole of CaBr2, 0.392 moles of CaBr2 have been removed from the solution. Similarly, each mole of the hydrate contains five moles of water, so the amount of water removed is:

Moles of H2O removed = 0.392 mol (CaBr2.5H2O) * 5 mol H2O/mol CaBr2.5H2O = 1.96 mol H2O

Now, let's convert the moles of water removed to grams:

Mass of H2O removed = 1.96 mol * 18.02 g/mol ≈ 35.32 g

Calculating Remaining CaBr2 and H2O

With the amount of precipitated CaBr2.5H2O accounted for, we can now determine the mass of CaBr2 and H2O remaining in the solution.

First, we find the mass of CaBr2 that precipitated:

Moles of CaBr2 precipitated = 0.392 mol (same as CaBr2.5H2O) Mass of CaBr2 precipitated = 0.392 mol * (40.08 + 2 * 79.90) g/mol = 0.392 mol * 199.88 g/mol ≈ 78.35 g

Now, we subtract this from the initial mass of CaBr2:

Mass of CaBr2 remaining = 200 g (initial) - 78.35 g (precipitated) ≈ 121.65 g

Similarly, we subtract the mass of water removed from the initial mass of water:

Mass of H2O remaining = 800 g (initial) - 35.32 g (precipitated) ≈ 764.68 g

Precipitation of Calcium Ions with Sodium Phosphate

The next stage involves precipitating the remaining calcium ions (Ca2+) in the solution using sodium phosphate (Na3PO4). The reaction between calcium ions and phosphate ions (PO4^3-) forms calcium phosphate (Ca3(PO4)2), an insoluble compound that precipitates out of the solution. The balanced chemical equation for this reaction is:

3Ca2+ (aq) + 2PO4^3- (aq) → Ca3(PO4)2 (s)

From the mass of CaBr2 remaining, we can calculate the moles of Ca2+ ions present:

Moles of CaBr2 remaining = 121.65 g / 199.88 g/mol ≈ 0.6086 mol

Since each mole of CaBr2 dissociates into one mole of Ca2+ ions, we have 0.6086 moles of Ca2+.

From the stoichiometry of the reaction, 3 moles of Ca2+ react with 2 moles of PO4^3-. Therefore, the moles of PO4^3- required to precipitate all Ca2+ ions are:

Moles of PO4^3- required = (2/3) * 0.6086 mol ≈ 0.4057 mol

Now, let's calculate the molar mass of Na3PO4:

Molar mass of Na3PO4 = 3 * 22.99 (Na) + 30.97 (P) + 4 * 16.00 (O) = 163.94 g/mol

The mass of Na3PO4 required is:

Mass of Na3PO4 required = 0.4057 mol * 163.94 g/mol ≈ 66.51 g

Determining the Mass of 10% Sodium Phosphate Solution

We need to determine the mass of a 10% sodium phosphate solution that contains 66.51 g of Na3PO4. A 10% solution means that 10 g of Na3PO4 are present in 100 g of solution. We can use this ratio to find the required mass of the solution:

Mass of 10% Na3PO4 solution = (Mass of Na3PO4 required / Percentage of Na3PO4) * 100 Mass of 10% Na3PO4 solution = (66.51 g / 10) * 100 ≈ 665.1 g

Therefore, approximately 665.1 grams of a 10% sodium phosphate solution are needed to precipitate the calcium ions remaining in the solution.

Conclusion

This problem exemplifies the intricate calculations involved in solution chemistry. We successfully navigated through the complexities of solubility, precipitation, and stoichiometry to determine the precise amount of sodium phosphate solution required to precipitate the remaining calcium ions. By meticulously analyzing each step, we've demonstrated a comprehensive understanding of the underlying chemical principles.

"Rewritten Keywords: Calculating sodium phosphate needed, calcium bromide solution cooling, phosphate precipitation, determining Na3PO4 solution mass, calcium ions precipitation"