Question

If a 25.0 mL water sample requires 27.2 mL of 0.104 M AgNO3 in such a titration, what is the

concentration of Cl- in the sample?

Answer 1

The balanced chemical equation for this titration is:

AgNO3 + Cl- → AgCl + NO3-

From the equation, we can see that 1 mole of AgNO3 reacts with 1 mole of Cl-. Therefore, the number of moles of Cl- in the water sample is equal to the number of moles of AgNO3 used in the titration.

Number of moles of AgNO3 = concentration × volume = 0.104 M × 27.2 mL = 2.8448 × 10^-3 moles

Since the volume of the water sample is 25.0 mL or 0.0250 L, the concentration of Cl- in the sample is:

Concentration of Cl- = number of moles / volume = 2.8448 × 10^-3 moles / 0.0250 L = 0.1138 M

Therefore, the concentration of Cl- in the water sample is 0.1138 M.