Question

You have combined the following: 7.65mL of 1.181M Fe(NO3)3(aq), 5.97mL of 2.158M NaCl(aq), and 7.24mL of water. What is the concentration of chloride ions in the resulting solution? Assume liquid volumes are additive and report your answer to 3 decimal places.

Answer 1

We mixed 7.65mL of 1.181M Fe(NO3)3(aq), 5.97mL of 2.158M NaCl(aq), and 7.24mL of water.

Hence total volume of the resulting solution after addition = 7.65mL+5.97mL+7.24mL = 20.86 mL.

When they are mixed Fe(NO₃)₃ will react with NaCl to form NaNO₃ and FeCl₃, but as both are soluble in water,hence the total Cl^- ion the solution is due to NaCl only.

Moles of NaCl in the solution = MxV(L) = 2.158 molL^-1x5.97x10^-3L = 2.158x5.97x10^-3 mol

Since 1 NaCl molecule contains 1 Cl^- ion,

hence moles of Cl^- ion in the solution = moles of NaCl in the solution = 2.158x5.97x10^-3 mol

$\therefore$ concentration of Cl^- in the solution = n/total volume(L) = 2.158x5.97x10^-3 mol/20.86x10^-3 L = 0.618 molL^-1 =0.618 M (Answer)