What is the concentration (in M) of NaCl in a solution if titration of 15.00 mL of the solution with 0.2255 M AgNO3 requires 19.30 mL of the AgNO3 solution to reach the end point?

Question

AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq)

Answer 1

NaCl(aq) + AgNO₃(aq) ==> AgCl(s) + NaNO₃(aq) ... balanced equation

First, find moles of AgNO3 needed:

19.30 ml x 1 L/1000 ml x 0.2255 moles/L = 0.004352 moles AgNO3

Next, find the moles of NaCl that will react with this many moles of AgNO3:

0.004352 moles AgNO3 x 1 mole NaCl/1 mole AgNO3 = 0.004352 moles NaCl

Now that we know the moles of NaCl that are present, we can easily calculate the concentration of NaCl:

The volume of NaCl used = 15.00 mls = 0.01500 L, so concentration is moles/L or 0.004352 moles/0.01500 L

[NaCl] = 0.004352 mol/0.01500 L = 0.29014 moles/L = 0.2901 M (to 4 significant figures)