Question

25.0ml of a 0.2450M NH4CL solution is added to 55.5ml of 0.1655M FeCL3. What is the concentration of chloride ion in the final solution?

Answer 1

Solution :-

lets first calculate the moles of the Cl- in each solutiuon

moles * molarity * volume in liter

moles of NH4Cl = 0.2450 mol per L * 0.025 L = 0.006125 mol NH4Cl = Cl-

moles of FeCl3 = 0.1655 mol per L * 0.0555 L = 0.00919 mol FeCl3

therefore

0.00919 mol FeCl3* 3 mol Cl- / 1 mol FeCl3 = 0.02757 mol Cl-

therefore total moles of Cl- = 0.006125 mol + 0.02757 mol = 0.0337 mol Cl-

total volume = 25.0 ml + 55.5 ml = 80.5 ml = 0.0805 L

now lets calculate the molarity of the Cl^-

[Cl^- ] = moles / volume in liter

         = 0.0337 mol / 0.0805 L

         = 0.419 M