To find the final concentration of phosphate anion (PO4^-3) in both ppm and molarity, we can follow these steps.
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Calculate the amount of PO4^-3 from each component before mixing:
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From the 0.5M orthophosphate solution:
- Volume = 50 mL = 0.050 L
- Concentration = 0.5 M
- Moles of PO4^-3 = Concentration × Volume = 0.5 mol/L × 0.050 L = 0.025 moles
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From the 500 ppm PO4^-3 solution:
- 500 ppm = 500 mg/L
- Volume = 35 mL = 0.035 L
- Concentration in mg/L can be converted to moles:
- Molar mass of PO4^-3 (Phosphate) is approximately 94.97 g/mol, or 94,970 mg/mol.
- Moles of PO4^-3 = (500 mg/L × 0.035 L) / 94,970 mg/mol
- Moles of PO4^-3 from 35 mL = (17.5 mg) / (94,970 mg/mol) = 0.000184 moles (approximately).
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Total moles of PO4^-3:
- Total moles = Moles from 0.5M solution + Moles from 500 ppm solution
- Total moles = 0.025 moles + 0.000184 moles = 0.025184 moles (approximately).
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Find the final volume of the solution:
- The final total volume of the solution is 250 mL = 0.250 L.
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Calculate the final molarity (polarity): \[ \text{Molarity (M)} = \frac{\text{Total moles}}{\text{Total volume in liters}} = \frac{0.025184 \text{ moles}}{0.250 \text{ L}} = 0.100736 \text{ M} \]
(To 7 significant figures: 0.1007360 M)
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Calculate the final concentration in ppm:
- We need to convert the total moles of PO4^-3 back to mg. \[ \text{mass of PO4} = \text{Total moles} \times \text{Molar mass} = 0.025184 \text{ moles} \times 94.97 \text{ g/mol} = 2.393 \text{ g} = 2393 \text{ mg} \]
Now, to find the concentration in ppm: \[ \text{ppm} = \frac{\text{mass of solute (mg)}}{\text{volume of solution (L)}} = \frac{2393 \text{ mg}}{0.250 \text{ L}} = 9572 \text{ ppm} \]
(To 4 significant figures: 9572 ppm)
Final Results:
- Concentration in ppm: 9572 ppm
- Concentration in molarity: 0.1007360 M
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