Question

75.0 mL of 0.200 M AgNO3 is combined with 125.0 mL of 0.100 M Nacl. solid AgCl forms.

a) what mass of AgCl forms ?

b) what is the concentration of each of the 4 ions in the final solution ? assume that volumes are additive. note.... none of the concentration will be zero.'

-Ag , Na , NO3 and Cl

Answer 1

AgNO3 + NaCl = AgCl + NaNO3

75 ml 0.2 M AgNO3 = 75 x 0.2/1000 moles = 0.015 moles

125 ml 0.1 M NaCl = 125 x 0.1/1000 moles = 0.0125 moles

1 mole NaCl reacts with 1 mole AgNO3 to form 1 mole AgCl

0.0125 moles NaCl react with  0.0125 moles AgNO3 to form 0.0125 moles AgCl

(a) Mass of AgCl formed = (0.0125 x 143.32) g = 1.7915 g.

(b) Excess AgNO3 = (0.015 - 0.0125) moles = 0.0025 moles.

Total Volume = (75 + 125 ) ml = 200 ml

AgNO3 concentration = 0.0025 x 1000 /200 = 0.0125 M

Ag+ concentration = 0.0125 M

NaNO3 formed = 0.0125 moles

NaNO3 Concentration = 0.0125 x 1000/200 = 0.0625 M

Na+ ion Concentration =0.0625 M

NO3- concentration = 0.0125 M + 0.0625 M = 0.075 M