Question

1. If you started with 33.52g of silver nitrate and added that to 700 mL of a 0.25M solution of sodium phosphate
a.   Write the reaction
b.   What would be the grams of precipitate produced
c.   Calculate the final molarity of the phosphate ion
d.   Calculate the final molarity of the silver ion

Answer 1

1) a) 3A9 Noglaq) + NagpOy (aq) → Ag, Pou (s) + 3NANG (0) - 0.197mal.. - by Moles of AgNO₃ = 33.52g 169.87 glmel Moles of Na, Poy = 0.25M X 0.7L T = 0.175 mol I mole Na, Pou reacts with 3 me AgNO₃ 10.175 mol of Napoy reacts with :- 3 me AgNO₃ x 0.175 med Na, Poy. Imol Na Pou = 0.52 mel AgNO3 . But there is only 0.197 mel AgNO₃. Therefore AgNO3 will be the limiting reagent... 3 mol of AgNO₃ forms I mel Ag, Pou (Precipitate) 10. 197 med Ag No, forme : - Imel Ag, Poy x 0.197 md AgNO₃ 3 ml AgNO₃- = 0.0657 mel. Ag Pou. weight of Precipitate = 0.0657 ml x. 418.58 g malt = 27.5 g Ag poy.

c) Nagpoy present in excess. Amount left = Initial moles - Meles which is used. 3 mel AgNO₃ reacts with 1 mol Na3PO4. 10.197 mel AgNO₃ reacts with 0.197 mel. L = 0.0657 mol Napoy. 3 left = - 0.175 mel - 0.0657 me 0.1093 mel. molarity = 0.1093 mel xlooo ToomL = 0.156M. Molacity of Pou - (d) Molarity of Ag Poy = meles volume = 0.0657 me 0.7L. = 0.094 M. Aga poy y 3 Ag+ + Poys- мо\ахич | g = 3xo 039 м = 0.282 M.