A 450.0 mL sample of a 0.242 M solution of silver nitrate is mixed with 400.0 mL of 0.200 M calcium chloride. What is the concentration of Cl - in solution after the reaction is complete?

0.538 M of Cl - in solution

Explanation:

First we need the balanced reaction:

2AgNO3 + CaCl2 ⇒ Ca (NO3) 2 + 2AgCl

Then, we calculate mol of Ag in the reaction.

mol Ag = 450mL * (0.242 molAgNO3 / 1000mL) = 0.1089 mol AgNO3

Now, how many moles of CaCl2 we need to react with 0.1089 mol AgNO3? We take this information, from balanced reaction.

0.1089 mol AgNO3 * (1 mol CaCl2 / 2 mol AgNO3) = 0.05445 mol CaCl2

We just need 0.054 mol CaCl2, and the rest of CaCl2 remain in solution.

so,

0.08 mol CaCl2 - 0.05445 mol CaCl2 = 0.0256 mol CaCl2 in excess,

Now, how many moles of Cl - are in 0.0256 mol of CaCl2?

0.0256 mol CaCl2 * 2 mol Cl-/1mol CaCl2 = 0.0511 mol Cl-

Molarity Cl - = Moles Cl- / Volumen (L)

V (L) = (450 mL + 400 mL) * (1L / 1000 mL) = 0.950 L

M = 0.0511mol Cl- / 0.950 L = 0.538 M