Answer to Question #99370 in Chemistry for Jesel Jamig

Question

The phosphate in a 4.258-g sample of plant food was precipitated as
Ag3PO4
through the addition of 50.00 mL of 0.0820 mol L-1 AgNO3
:
3Ag+ + HPO42- ⇌ Ag3PO4(s) + H+
The solid was filtered and washed, following which the filtrate and
washings were diluted to exactly 250.0 mL. Titration of a 50.00-mL
aliquot of this solution required a 4.64-mL back-titration with 0.0625
mol L-1 KSCN. Express the results of this analysis in terms of the
percentage of P2O5
.

Answer 1

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