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

Answer:The first step is to calculate the moles of AgNO3 used:

0.0500 L x 0.082 mol/L = 0.0041 mol AgNO3

Since Ag3PO4 is formed in a 1:3 stoichiometric ratio with PO43-, the number of moles of PO43- is also 0.0041 mol.

The molar mass of Ag3PO4 is 418.6 g/mol:

0.0041 mol Ag3PO4 x 418.6 g/mol = 1.74 g Ag3PO4

Therefore, the percentage of P2O5 in the sample is:

(2 x 30.97 g/mol x 0.0041 mol / 4.258 g sample) x 100% = 30.0%

Note that the factor of 2 is used because P2O5 is the anhydride of H3PO4, and its molar mass is twice that of P.