By now, you know from the solubility rules and the solubility table that silver

Question

By now, you know from the solubility rules and the solubility table that silver chloride is very insoluble in water. Chemists know that by adding a very slight excess of silver nitrate to a solution containing chloride ion, it is possible to effectively remove all the chloride ion from the sample as silver chloride. First, write a balanced net ionic for reaction. Now, assume that a sample has a mass of 10,000 grams and requires 7.956 gm of silver nitrate to completely react with all the chloride in the sample. Calculate the percentage of soluble chloride ion in the sample.

Explanation / Answer

Cl-(aq) + Ag+(aq) +NO3-(aq) AgCl(s) +NO3-(aq)

nitrate ions cancels out and net ionic equation is

Cl-(aq) + Ag+(aq) AgCl(s)

Molecular weight of AgNO3 = 169.87 g/mol

Atomic weight of Ag = 107.8682 g/mol

169.87 g AgNO3 contains 107.8682 g Ag

So 7.956 g AgNO3 contains = (107.868 x 7.956)/169.87 = 5.052 g Ag

1 mole AgCl = 143.32 g/mol

107.868 g Ag combines with 35.453 g Cl

So 5.052 g Ag needs = (35.453 x5.052)/107.868 = 1.660 g Cl

So in 10.00 g water 1.660 g Cl is present.

% of chloride ion = (1.660 x100)/10.00 = 16.60 %

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