Question

A solution is made by mixing 75.0 ml of 0.350 M BaCl_2 with 30.0 ml of 0.800 M LiCl. What is the molar concentration of the chloride ion in the new solution?

Answer 1

BaCl2--->Ba2+ +2Cl-

LiCl--->Li+ +Cl-

Cl- is the common ion in the solution, but both LiCl and BaCl2 are quite soluble in water,so they ionizes almost completely.

calculation of molar concentration of Cl- due to solubility of BaCl2-

[BaCl2]=0.075L*0.350 mol/L=0.0262 moles

2[BaCl2]=[Cl-]=2*0.0262 moles=0.0525 moles

calculation of molar concentration of Cl- due to solubility of LiCl

[LiCl]=0.030L*0.8mol/L=0.024 moles

[LiCl]=[Cl-]=0.024 moles

total moles of Cl- in solution=0.0525+0.024=0.0765 moles

total volume of solution=75ml+30ml=105ml=0.105L

So,[Cl-]=moles of Cl-/total volume of the solution=0.0765moles/0.105L=0.728mol/L=0.728M