By convention (due to tradition), plant fertilizers report the amount of nitroge
ID: 825577 • Letter: B
Question
By convention (due to tradition), plant fertilizers report the amount of nitrogen,
phosphorus, and potassium (the N-P-K number) as the amount of nitrogen,
diphosphorus pentoxide, and potassium oxide present in 100 pounds of the fertilizer.
By reacting solid diphosphorus pentoxide with liquid water, phosphate ions and
hydronium ions are produced. The phosphate ions can then react with silver (I) nitrate
to form a precipitate, solid silver (I) phosphate. Finally, the excess silver (I) nitrate can
be back titrated with potassium thiocyanate, KSCN, to form solid silver (I)
thiocyanate.
A 57.84 mg sample of plant food was mixed with water. To this solution 42.76 mL of
0.0921 M silver (I) nitrate was added. Finally, the excess silver (I) nitrate was back
titrated with 1.850 mL of 0.0715 M potassium thiocyanate. Write the balanced
chemical equation for each reaction and determine the weight % of phosphorus in the
sample of plant food.
Explanation / Answer
The balanced chemical equations for the reactions are:
P2O5(s) + 9 H2O(l) => 6 H3O+(aq) + 2 PO43-(aq)
3 AgNO3(aq) + PO43-(aq) => Ag3PO4(s) + 3 NO3-(aq)
AgNO3(aq) + KSCN(aq) => AgSCN(s) + KNO3(aq)
Total moles of AgNO3 = volume x concentration of AgNO3
= 42.76/1000 x 0.0921 = 0.003938196 mol
Moles of excess AgNO3 = moles of KSCN = volume x concentration of KSCN
= 1.850/1000 x 0.0715 = 0.000132275 mol
Moles of reacted AgNO3 = total - excess moles of AgNO3
= 0.003938196 - 0.000132275 = 0.003805921 mol
Moles of P = moles of PO43- = 1/3 x moles of reacted AgNO3
= 1/3 x 0.003805921 = 0.00126864 mol
Mass of P = moles x molar mass of P
= 0.00126864 x 30.974 = 0.039295 g
Weight% of P = mass of P/mass of sample x 100%
= 0.039295/(57.84 x 10^(-3))
= 67.94% = 67.9%
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