Initial Knowledge Check 58WgZD-OpHQzMM8F One way the u.s. Environmental Protecti

Question

Initial Knowledge Check 58WgZD-OpHQzMM8F One way the u.s. Environmental Protection Agency (EPA) tests for chloride contaminants in water is by titrating a sample of silver nitrate solution Any chloride anions in solution will combine with the silver cations to produce bright white silver chloride precipitate Suppose an EPA chemist tests a 200. mL sample of groundwater known to be contaminated with nickel(!I) chloride, which would react with silver nitrate solution like this: Nicl2(aq) + 2 AgNO3(aq) 2 AgCl(s) + Ni(NO3)2(aa) The chemist adds 61.0 mM precipitate. He finds he has collected 5.6 mg of silver chloride. Calculate the concentration of nickel(I1) chloride contaminant in the original groundwater sample. Be sure your answer has the correct number of silver nitrate solution to the sample until silver chloride stops forming. He then washes, dries, and weighs the significant digits. 0% MacBook Pro

Explanation / Answer

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Ans. # Mass of AgCl precipitated = 5.6 mg = 0.0056 g             ; [1 mg = 0.001 g]

Moles of AgCl precipitated = Mass / Molar mass

                                                = 0.0056 g/ (143.3209 g/ mol)

                                                = 3.9073 x 10-5 mol

# Balanced reaction:             NiCl2(aq) + 2 AgNO3(aq) ----> 2 AgCl(s) + Ni(NO3)2(aq)

According to the stoichiometry of balanced reaction, 1 mol NiCl2 forms 2 mol AgCl.

So,

Total moles of NiCl2 consumed = (1 mol NiCl2 / 2 mol AgCl) x Moles of AgCl ppt.

                                                = (1 mol NiCl2 / 2 mol AgCl) x 3.9073 x 10-5 mol AgCl

                                                = 1.9537 x 10-5 mol NiCl2

Therefore, the 200.0 mL of ground water sample consists of 1.9537 x 10-5 mol NiCl2.

# We have-

                        Moles of NiCl2 = 1.9537 x 10-5 mol

                        Volume of groundwater sample = 200.0 mL = 0.200 L

Now,

            [NiCl2] = Moles of NiCl2/ Volume of sample in liters

                        = 1.9537 x 10-5 mol / 0.200 L

                        = 9.7655 x 10-5 M  

That is, there is 9.7655 x 10-5 mole NiCl2 in 1.0 L of sample water.

Now,

            Using-             mass = moles x molar mass

            [NiCl2] = (9.7655 x 10-5 mol x 129.5988 g mol-1) / L

            Or, [NiCl2] = 1.266 x 10-2 g / L

            Or, [NiCl2] = 1.266 x 10-2 X (103 mg) / L

            Hence, [NiCl2] = 1.3 x 101 mg/ L

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