The Ba3(PO4)2 (molar mass = 601.93 g/mol) precipitate that formed from a salt mixture has a mass of 0.667 g. Experimental tests revealed that Na3PO4•12H2O (molar mass = 380.12 g/mol) was the limiting reactant in the formation of the precipitate and the BaCl2•2H2O was the excess reactant in the salt mixture. Determine the mass of Na3PO4•12H3O in the salt mixture. Barium phosphate is a very finely divided precipitate. Identify two special steps in the procedure that are incorporated into the experiment that minimize the loss of barium phosphate in its analysis.
The Correct Answer and Explanation is :
Solution:
1. Determine the moles of Ba3(PO4)2\text{Ba}_3(\text{PO}_4)_2:
The mass of Ba3(PO4)2\text{Ba}_3(\text{PO}_4)_2 precipitate is 0.667 g, and its molar mass is 601.93 g/mol. Moles of Ba3(PO4)2=MassMolar Mass=0.667601.93=0.001108 mol.\text{Moles of } \text{Ba}_3(\text{PO}_4)_2 = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{0.667}{601.93} = 0.001108 \text{ mol.}
2. Reaction Stoichiometry:
The balanced chemical equation is: 2Na3PO4+3BaCl2→Ba3(PO4)2+6NaCl.2 \text{Na}_3\text{PO}_4 + 3 \text{BaCl}_2 \rightarrow \text{Ba}_3(\text{PO}_4)_2 + 6 \text{NaCl.}
From the stoichiometry, 2 moles of Na3PO4\text{Na}_3\text{PO}_4 produce 1 mole of Ba3(PO4)2\text{Ba}_3(\text{PO}_4)_2. Therefore: Moles of Na3PO4=2×Moles of Ba3(PO4)2=2×0.001108=0.002216 mol.\text{Moles of } \text{Na}_3\text{PO}_4 = 2 \times \text{Moles of } \text{Ba}_3(\text{PO}_4)_2 = 2 \times 0.001108 = 0.002216 \text{ mol.}
3. Mass of Na3PO4⋅12H2O\text{Na}_3\text{PO}_4 \cdot 12\text{H}_2\text{O}:
The molar mass of Na3PO4⋅12H2O\text{Na}_3\text{PO}_4 \cdot 12\text{H}_2\text{O} is 380.12 g/mol. The mass is calculated as: Mass=Moles×Molar Mass=0.002216×380.12=0.842 g.\text{Mass} = \text{Moles} \times \text{Molar Mass} = 0.002216 \times 380.12 = 0.842 \text{ g.}
Thus, the mass of Na3PO4⋅12H2O\text{Na}_3\text{PO}_4 \cdot 12\text{H}_2\text{O} in the salt mixture is 0.842 g.
Special Steps to Minimize Loss of Ba3(PO4)2\text{Ba}_3(\text{PO}_4)_2:
- Washing the Precipitate with Dilute Electrolyte Solution: Washing the precipitate with a dilute electrolyte, such as HNO3\text{HNO}_3, minimizes the loss due to dissolution during rinsing. Water alone could cause re-dissolution.
- Use of a Gooch Crucible or Filter Paper: Filtering the finely divided precipitate using a Gooch crucible or fine-grade filter paper ensures that the tiny particles are not lost during filtration.
These precautions are critical because Ba3(PO4)2\text{Ba}_3(\text{PO}_4)_2 is finely divided and prone to losses during transfer and washing, which could lead to errors in mass measurements.