An aqueous solution contains \(1.2 \mathrm{~m} M\) of total ions. (a) If the solution is \(\mathrm{NaCl}(a q),\) what is the concentration of chloride ion? (b) If the solution is \(\mathrm{FeCl}_{3}(a q),\) what is the concentration of chloride ion? [Section 4.5\(]\)

In NaCl(aq) solution, sodium chloride dissociates into its constituent ions: sodium ions (Na⁺) and chloride ions (Cl⁻). The dissociation of one mole of NaCl produces one mole of Na⁺ and one mole of Cl⁻ ions, so the ratio between NaCl and Cl⁻ is 1:1. Given the total ion concentration in the solution is 1.2 M, we can write this as the sum of the concentrations of Na⁺ and Cl⁻ ions: \[1.2 ~\text{M} = [Na^+] + [Cl^-]\] Since the ratio between NaCl and Cl⁻ is 1:1, their concentrations are equal: \[[Na^+] = [Cl^-]\] Substituting this back into the equation, we can determine the concentration of chloride ions: \[1.2 ~\text{M} = 2[Cl^-]\] Now, solve for [Cl⁻]: \[[Cl^-] = \frac{1.2 ~\text{M}}{2} = 0.6 ~\text{M}\] So, the concentration of chloride ions in the NaCl(aq) solution is 0.6 M.

In FeCl₃(aq) solution, iron(III) chloride dissociates into its constituent ions: iron(III) ions (Fe³⁺) and chloride ions (Cl⁻). The dissociation of one mole of FeCl₃ produces one mole of Fe³⁺ and three moles of Cl⁻ ions, so the ratio between FeCl₃ and Cl⁻ is 1:3. Given the total ion concentration in the solution is 1.2 M, we can write this as the sum of the concentrations of Fe³⁺ and Cl⁻ ions: \[1.2 ~\text{M} = [Fe^{3+}] + 3[Cl^-]\] Since the ratio between FeCl₃ and Cl⁻ is 1:3, their concentrations are related as: \[[Fe^{3+}] = \frac{1}{3}[Cl^-]\] Substituting this back into the equation, we can determine the concentration of chloride ions: \[1.2 ~\text{M} = \frac{1}{3}[Cl^-] + 3[Cl^-]\] Combine the Cl⁻ terms: \[1.2 ~\text{M} = \frac{10}{3}[Cl^-]\] Now, solve for [Cl⁻]: \[[Cl^-] = \frac{1.2 ~\text{M}}{\frac{10}{3}} = \frac{1.2 ~\text{M} \cdot 3}{10} = 0.36 ~\text{M}\] So, the concentration of chloride ions in the FeCl₃(aq) solution is 0.36 M.