To find the final concentration of phosphate anions (\( \text{PO}_4^{3-} \)) in both ppm and molarity, we first need to calculate the amount of phosphate contributed from each solution and then determine the final volume of the mixed solution.
Step 1: Calculate moles of phosphate from the two solutions.
1. From the 0.5 M orthophosphate solution:
- Volume = 50 mL = 0.050 L
- Molarity (M) = 0.5 M
\[ \text{Moles of PO}_4^{3-} = \text{Molarity} \times \text{Volume} = 0.5 , \text{mol/L} \times 0.050 , \text{L} = 0.025 , \text{mol} \]
2. From the 5000 ppm phosphate solution:
- 5000 ppm means 5000 mg of \( \text{PO}_4^{3-} \) in 1 L.
To convert this to moles:
-
Molar mass of \( \text{PO}_4^{3-} \) \( ( \text{P} = 30.97 , \text{g/mol}, \text{O} = 16.00 , \text{g/mol}) \) \[ \text{Molar mass of PO}_4^{3-} = 30.97 + 4 \times 16.00 = 94.97 , \text{g/mol} \]
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Concentration in mg/L: \[ \text{Molarity (M)} = \frac{5000 , \text{mg/L}}{94.97 , \text{g/mol}} = \frac{5000 , \text{mg/L}}{94.97 , \text{g/mol}} \times \frac{1 , \text{g}}{1000 , \text{mg}} = 0.0527 , \text{mol/L} \]
Now calculate the moles in 35 mL:
- Volume = 35 mL = 0.035 L \[ \text{Moles of PO}_4^{3-} = 0.0527 , \text{mol/L} \times 0.035 , \text{L} = 0.00184 , \text{mol} \]
Step 2: Total moles of \( \text{PO}_4^{3-} \)
Adding the moles from both solutions: \[ \text{Total moles of PO}_4^{3-} = 0.025 , \text{mol} + 0.00184 , \text{mol} = 0.02684 , \text{mol} \]
Step 3: Calculate final concentration in molarity
Total volume of the mixed solution is 250 mL = 0.250 L. \[ \text{Final Molarity} = \frac{\text{Total moles}}{\text{Total volume}} = \frac{0.02684 , \text{mol}}{0.250 , \text{L}} = 0.10736 , \text{mol/L} \]
Step 4: Calculate final concentration in ppm
To convert to ppm (mg/L): \[ \text{Total mass of PO}_4^{3-} = \text{Total moles} \times \text{Molar mass} = 0.02684 , \text{mol} \times 94.97 , \text{g/mol} = 2.5487 , \text{g} \] Convert to mg: \[ 2.5487 , \text{g} = 2548.7 , \text{mg} \]
Now, since the total volume is 250 mL: \[ \text{Final Concentration in ppm} = \frac{2548.7 , \text{mg}}{0.250 , \text{L}} = 10194.8 , \text{ppm} \]
Final Results
Concentration in ppm to four significant figures: \[ \text{Final concentration in ppm: } 10190 , \text{ppm} \] Concentration in molarity to seven significant figures: \[ \text{Final concentration in molarity: } 0.1073600 , \text{mol/L} \]
So the final concentrations are:
- \( 10190 , \text{ppm} \)
- \( 0.1073600 , \text{mol/L} \)