Can someone explain in detail how to do this problem? If volumes are additive and 125.0 mL of 0.125M NaCl is mixedwith 200 mL of an Iron (III) Chloride solution to give a newsolution in which [Cl-] is 0.509 M , what is the concentration ofthe Iron (III) Chloride used to make the new solution? Can someone explain in detail how to do this problem? If volumes are additive and 125.0 mL of 0.125M NaCl is mixedwith 200 mL of an Iron (III) Chloride solution to give a newsolution in which [Cl-] is 0.509 M , what is the concentration ofthe Iron (III) Chloride used to make the new solution?
Explanation / Answer
We Know that : FeCl3 + NaCl* -----> FeCl2Cl* + NaCl When the twosolutions are mixed the Resulting solution has the molarity as: M = M1V1 + M2V2 / ( V1 + V2 ) 0.509 M = 0.125 M * 125 ml + MFeCl3 * 200 ml / (125 ml + 200 ml ) MFeCl3 = 0.749 M of FeCl3 has taken so that it will results theconcentration of Cl- is 0.509 M .