Question

Addition of AgNO₃ to aqueous solutions of the complex results in a cloudy white precipitate, presumably AgCl. You dissolve 0.1000 g of the complex in H₂O and perform a precipitation titration with 0.0500 M AgNO₃ as the titrant. Using an electrode that is sensitive to [Ag⁺], you reach the endpoint after 9.00 mL of titrant are added. How many grams of chloride ion were present in the 0.1000-g sample?

(The answer is E, Could anyone explain it ?

A)	4.50 ´ 10–4
B)	5.00 ´ 10–3
C)	1.77 ´ 10–3
D)	6.38 ´ 10–2
E)	1.60 ´ 10–2

Answer 1

Reaction occurs:

Ag⁺ (aq) + Cl^- (aq) = AgCl (s)

According to above equation:

n(Ag⁺ ) = n(Cl^-)

n-moles

From titration:

n=c*V

c-concentration, mol/L

V-volume, L

n(Ag⁺ ) = n(AgNO₃) = 0.05 mol/L * 0.009 L (Volume added to endpoint in Liters) = 0.00045 mol = 4.5*10^-4 moles

As mentioned above:

n(Cl^-) = n(Ag⁺ ) = n(AgNO₃) = 4.5*10^-4 moles

m = n*M

m-mass, in g

M-molar mass in g/mol, for anion Cl^- 35.45 g/mol. So

m(Cl^-) = 4.5*10^-4 moles * 35.45 g/mol = 1.595*10^-2 g = 1.6*10^-2 g

Answer - E)