A 450.0 mL sample of a 0.268M 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?
-
2 AgNO3 + CaCl2 → 2 AgCl(s) + Ca(NO3)2
(450.0 mL) x (0.268 M AgNO3) = 120.6 mmol AgNO3
(400.0 mL) x (0.200 M CaCl2) = 80.0 mmol CaCl2
120.6 mmol AgNO3 would react completely with 120.6 x (1/2) = 60.3 mmol CaCl2 but there is more CaCl2 present than that, so CaCl2 is in excess.
(80.0 mmol CaCl2 initially) - (60.3 mmol CaCl2 reacted and precipitated) = 19.7 mmol CaCl2 remaining in solution
(19.7 mmol CaCl2) x (2 mol Cl{-} / 1 mol CaCl2) = 39.4 mmol Cl{-}
Supposing additive volumes:
(39.4 mmol) / (450.0 mL + 400.0 mL) = 0.0464 mol/L Cl{-}
(450.0 mL) x (0.268 M AgNO3) = 120.6 mmol AgNO3
(400.0 mL) x (0.200 M CaCl2) = 80.0 mmol CaCl2
120.6 mmol AgNO3 would react completely with 120.6 x (1/2) = 60.3 mmol CaCl2 but there is more CaCl2 present than that, so CaCl2 is in excess.
(80.0 mmol CaCl2 initially) - (60.3 mmol CaCl2 reacted and precipitated) = 19.7 mmol CaCl2 remaining in solution
(19.7 mmol CaCl2) x (2 mol Cl{-} / 1 mol CaCl2) = 39.4 mmol Cl{-}
Supposing additive volumes:
(39.4 mmol) / (450.0 mL + 400.0 mL) = 0.0464 mol/L Cl{-}