Suppose you are studying coordination compounds of Co(II) with the ligand pyridine (py, C5H5N, MW = 79.10). You isolate a crystalline compound and since the only available ions are chloride and nitrate, you hypothesize the empirical formula of the coordinaton compound must be Co(II)w(py)x(Cl)y(NO3)z. 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 water and perform a precipitation titration with 0.0500 M AgNO3 at the titrant. Using an electrode that is sensive 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?
5.00 x 10-3
4.50 x 10-4
1.60 x 10-2
6.38 x 10-2
1.77 x 10-3
L x M x atomic mass Cl = g Chloride.
To solve this problem, we can use stoichiometry and titration concepts.
First, let's calculate the number of moles of AgNO3 used in the titration:
Moles of AgNO3 = Concentration (M) x Volume (L)
Moles of AgNO3 = 0.0500 M x 9.00 mL = 0.4500 mmol
Next, we need to determine the stoichiometry between AgNO3 and chloride ions (Cl-) in the complex. From the given information, it is stated that the cloudy white precipitate formed is AgCl. This indicates that 1 mole of AgNO3 reacts with 1 mole of Cl- to form 1 mole of AgCl.
Now, we can calculate the number of moles of chloride ions in the complex:
Moles of Cl- = Moles of AgNO3
Moles of Cl- = 0.4500 mmol
Finally, we can convert the moles of chloride ions to grams of chloride:
Grams of chloride = Moles of Cl- x Atomic mass of Cl
Grams of chloride = (0.4500 mmol / 1000) x 35.45 g/mol (atomic mass of Cl)
Grams of chloride = 0.0159 g
Therefore, the correct answer to the question is 1.60 x 10-2 g (rounded to three significant figures).