Question

Suppose an EPA chemist tests a 250.0 ml sample of groundwater known to be contaminated with nickel(II) chloride, which would react with silver nitrate solution like this:
NiCl_2(aq) + 2AgNO_3 (aq) -------> AgCl (s) + Ni(NO_3)_2 (aq)
The chemist adds 58.0m silver nitrate solution to the sample until silver chloride stops forming. He then washes, dries, and weighs the precipitate. He finds he has collected of 3.6 mg of silver chloride.
1. Calculate the concentration of nickel(II) chloride contaminant in the original groundwater sample. Round your answer to 2 significant digits.

Answer 1

Answer:

0.05 mM

Explanation:

NiCl_2(aq) + 2AgNO_3 (aq) -------> 2 AgCl (s) + Ni(NO_3)_2 (aq)

Molar mass of  NiCl_2 = 129.5994 g/mol

Molar mass of AgNO_3 = 169.87 g/mol

Molar mass of AgCl = 143.32 g/mol

The chemist added 58mM of silver nitrate = 58 / 1000 = 0.058 M

mass of silver nitrate present =0.058 × 169.87 = 9.85246 g

according to the equation of reaction

2 mole of silver nitrate is needed to produce 2 mole of silver chloride  

2 mole of silver nitrate = 339.74 g

2 mole of silver chloride = 286.64 g

339.74 g silver nitrate  produces 286.64 g silver chloride

unknown mass of silver nitrate produce 0.0036 g

unknown mass of silver nitrate = 339.74 × 0.0036 / 286.64 = 0.00427 g

then

1 mole of NiCl_2 reacts 2 mole of AgNO_3

129.5994 g of NiCl_2 react with 339.74 g AgNO_3

unknown mass of NiCl_2  react with  0.00427 g AgNO_3

mass of NiCl_2 = 0.00163 g

mole of NiCl_2 present = 0.00163 / 129.5994 = 0.0000126 mole

Molarity = 0.0000126 mole / 0.25 L = 0.0000504 M = 0.05 mM