Suppose you are studying coordination compounds of Co(II) with the ligand pyridine (py, C5H5N, molar mass=79.10). You isolate a crystalline compound, and because the only available anions are Cl- and NO3-, you hypothesize that the empirical formula of the coordination compound must be Cow(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 0.100 g of the complex in H2O and perform a precipitation titration with 0.0500 M AgNO3 as the titrant. Using an electrode that is sensitive to [Ag+], you reach the endpoint after 9.00 mL of titrant is added. How many grams of chloride ion were present in the 0.100g sample?

To determine the grams of chloride ion present in the 0.100 g sample, we need to use the information provided in the question.

First, let's analyze the reaction that takes place during the precipitation titration between the complex and AgNO3. From the observation of a white precipitate, we can infer that AgCl is formed:

Co(py)x(Cl)y(NO3)z + AgNO3 → AgCl + Co(py)x(NO3)z

From this equation, we can see that each mole of AgNO3 reacts with one mole of chloride ion (Cl-) to form one mole of AgCl.

Given that the concentration of the AgNO3 titrant is 0.0500 M and the volume at the endpoint is 9.00 mL (which can be converted to liters by dividing by 1000), we can calculate the moles of AgNO3 used:

moles of AgNO3 = concentration × volume
= 0.0500 M × (9.00 mL / 1000)
= 0.00045 mol

Since the mole ratio of AgNO3 to chloride ions is 1:1, the number of moles of chloride ions is also 0.00045 mol.

To determine the mass of chloride ions, we need to use the molar mass of chloride (Cl-). The molar mass of chloride is equal to the molar mass of chlorine, which is 35.45 g/mol.

mass of chloride ions = moles of chloride ions × molar mass of chloride
= 0.00045 mol × 35.45 g/mol
= 0.016 g

Therefore, the 0.100 g sample of the complex contained 0.016 g of chloride ion.

To determine the number of grams of chloride ion present in the 0.100 g sample, we need to use the information from the precipitation titration.

1. Find the number of moles of Ag+ ions used in the titration:
Given: Volume of titrant = 9.00 mL = 0.00900 L
Concentration of AgNO3 = 0.0500 M

Number of moles of Ag+ ions used = concentration × volume
= 0.0500 M × 0.00900 L

2. Determine the stoichiometry of the reaction between Ag+ ions and Cl- ions:
From the information provided, it is known that AgNO3 reacts with Cl- ions to form a precipitate of AgCl. The balanced equation is:
Ag+ (aq) + Cl- (aq) → AgCl (s)

According to the stoichiometry, each Ag+ ion reacts with one Cl- ion to form one mole of AgCl.

3. Convert the moles of Ag+ ions used to moles of Cl- ions:
Since one mole of Ag+ ions reacts with one mole of Cl- ions, the number of moles of Cl- ions will be the same as the number of moles of Ag+ ions used in the titration.

4. Convert the moles of Cl- ions to grams of chloride ion:
To convert moles of Cl- ions to grams of chloride ion, we need to know the molar mass of chloride ion, Cl-.

The molar mass of Cl- is 35.45 g/mol.

Mass of chloride ion = Number of moles × molar mass
= (0.0500 M × 0.00900 L) × 35.45 g/mol

Now, you can calculate the mass of chloride ion present in the 0.100 g sample.

All of that extraneous information makes one think this is a tough problem. Not so.

mols Ag present = M x L = ?
grams = mols Ag x atomic mass Ag.
By the way you must assume with this that the Cl ion is not "within" the complex but outside the coordination sphere so that it acts as a Cl ion.