Question

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 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 0.1000-g sample

Answer 1

0.016 grams of chloride ion were present in the 0.1000 grams of sample.

Step-by-step explanation:

According to question, 9.00 mL of titrant was added to solution with 0.1000 grams of complex to reach the end point.

Molarity of the silver nitrate solution = 0.0500 M

Volume of the silver nitrate solution = V = 9.00 mL = 0.009 L

1 mL = 1000 L

Moles of silver nitrate = n

`Molarity=(Moles)/(Volume (L))`

`0.0500 M=(n)/(0.009 L)`

n = 0.00045 mol

`Cl^-+AgNO_3\rightarrow AgCl+NO_3^(-)`

According to 1 mole of silver nitrate reacts with 1 mol of chloride ion, then 0.00045 moles of silver nitrate will :

`(1)/(1)* 0.00045 mol=0.00045 mol` of chloride ions

Mass of chloride ions :

0.00045 mol × 35.5 g/mol = 0.016 g

0.016 grams of chloride ion were present in the 0.1000 grams of sample.