The quantity of Cl- in a water supply is determined by titrating the sample with Ag+. What mass of chloride ion is present in a 10,0g sample of the water if 20.2 ml of 0.100 M Ag+ is required to react with all the chloride in the sample?
Cl^- + Ag^+ ==> AgCl
mols Ag^+ = M x L = ?
Because of the 1:1 ratio
mols Cl^- = mols Ag^+.
Then g Cl^- = mols Cl- x atomic mass Cl-
To find the mass of chloride ions present in the water sample, we can use the concept of stoichiometry.
First, let's write down the balanced chemical equation for the reaction between chloride ions (Cl-) and silver ions (Ag+):
Ag+ + Cl- → AgCl
From the equation, we can see that 1 mole of Ag+ reacts with 1 mole of Cl-.
Next, let's determine the number of moles of Ag+ used in the titration. We can use the formula:
moles = concentration (M) x volume (L)
Given that the volume is 20.2 ml and the concentration is 0.100 M, we have:
moles of Ag+ = 0.100 M x 0.0202 L = 0.00202 moles
Since the reaction is 1:1 between Ag+ and Cl-, we can conclude that there are also 0.00202 moles of Cl- in the sample.
Now, we can calculate the molar mass of chloride ions (Cl-), which is 35.45 g/mol.
Finally, we can find the mass of chloride ions in the water sample using the formula:
mass = moles x molar mass
mass of Cl- = 0.00202 moles x 35.45 g/mol = 0.0716 g
Therefore, the mass of chloride ions present in the 10.0 g water sample is approximately 0.0716 grams.