Question

What mass in grams of a 10.0% by mass aqueous silver nitrate solution must be added to 100.0 mL of an aqueous solution that is 0.0250 M in chromium (II) chloride and 0.0200 M in chromium (III) chloride to precipitate all chloride ion as silver chloride from solution?

Answer 1

Reaction with 0.0250 M chromium (II) chloride:
CrCl₂(aq) + 2AgNO₃(aq) ------>    2AgCl(s)    +    Cr(NO₃)₂(aq)

1 mole of CrCl₂ reacts with 2 moles of AgNO₃ to precipitate 2 moles of AgCl(s).

no. moles of CrCl₂ = volume in L x molarity = 0.10L x 0.0250 = 0.0025 moles
Therefore, no. of moles of AgNO₃ required = 2 x 0.0025 moles = 0.05 moles

Mass of AgNO₃ = no. moles x mol.wt = 0.05 moles x 170 g/mol
                                   = 8.5 g.
10% AgNO₃ solution means, in 100g of a solution, we have 10 g of AgNO₃ and 90 g water.
10 g AgNO₃ in ----100 g solution,
for 8.5 g --------?
Mass of 10% AgNO₃ solution = 8.5g x 100 g/10g = 85 g

Therefore, we need 85 g of 10% AgNO₃ solution to precipitate all chloride ions as AgCl from the 0.0250 M chromium (II) chloride solution.

Reaction with 0.020 M chromium (III) chloride:
CrCl₃(aq) + 3AgNO₃(aq) ------>    3AgCl(s)    +    Cr(NO₃)₃(aq)

1 mole of CrCl₂ reacts with 3 moles of AgNO₃ to precipitate 3 moles of AgCl(s).

no. moles of CrCl2 = volume in L x molarity = 0.10L x 0.020 = 0.002 moles
Therefore, no. of moles of AgNO₃ required = 3 x 0.002 moles = 0.06 moles

Mass of AgNO₃ = no. moles x mol.wt = 0.06 moles x 170 g/mol
                                   = 10.2 g.
10% AgNO₃ solution means, in 100g of a solution, we have 10 g of AgNO₃ and 90 g water.
10 g AgNo₃ in ----100 g solution,
for 10.2 g --------?
Mass of 10% AgNO₃ solution = 10.2g x 100 g/10g = 102 g

Therefore, we need 102 g of 10% AgNO₃ solution to precipitate all chloride ions as AgCl from the 0.0200 M chromium (III) chloride solution.