Question

The phosphorus in a 4.258 g sample of a plant food was converted to PO43- and precipitated as Ag3PO4 through the addition of 50.0 mL of 0.082 M AgNO3. The excess of AgNO3 was back-titrated with 4.06 mL of 0.0625 M KSCN. Express the results of this analysis in terms of %P2O5 (141.94 g/mol).

Answer 1

The analysis indicates that the plant food sample contains approximately 13.63% P2O5.

Step-by-step explanation:

To calculate the percentage of P2O5 in the plant food sample, we need to determine the amount of phosphorus (P) present in the sample and then convert it to P2O5.

1. Calculate the moles of AgNO3 used in the precipitation reaction:

• Moles of AgNO3 = Volume of AgNO3 (in L) × Concentration of AgNO3 (in mol/L)

• Moles of AgNO3 = 0.0500 L × 0.082 mol/L = 0.0041 mol

2. Since the stoichiometric ratio between AgNO3 and Ag3PO4 is 3:1, the moles of Ag3PO4 formed in the precipitation reaction are also 0.0041 mol.

3. Calculate the moles of phosphorus (P) in the Ag3PO4:

Moles of P = Moles of Ag3PO4 × (1 mol of P / 1 mol of Ag3PO4)

Moles of P = 0.0041 mol × 1 mol / 1 mol = 0.0041 mol

4. Calculate the mass of P2O5:

Mass of P2O5 = Moles of P × Molar mass of P2O5

Mass of P2O5 = 0.0041 mol × 141.94 g/mol = 0.581 g

5. Calculate the percentage of P2O5 in the plant food sample:

%P2O5 = (Mass of P2O5 / Mass of the sample) × 100

%P2O5 = (0.581 g / 4.258 g) × 100 = 13.63%