The solubility of \(\mathrm{Cl}_{2}(\mathrm{g})\) in water is \(6.4 \mathrm{g} \mathrm{L}^{-1}\) at \(25^{\circ} \mathrm{C}\) Some of this chlorine is present as \(\mathrm{Cl}_{2},\) and some is found as HOCl or Cl^- For the hydrolysis reaction $$\begin{array}{c}\mathrm{Cl}_{2}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(1) \longrightarrow \\ \mathrm{HOCl}(\mathrm{aq})+\mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq}) \\\K_{c}=4.4 \times 10^{-4}\end{array}$$ For a saturated solution of \(\mathrm{Cl}_{2}\) in water, calculate \(\left[\mathrm{Cl}_{2}\right],[\mathrm{HOCl}],\left[\mathrm{H}_{3} \mathrm{O}^{+}\right],\) and \(\left[\mathrm{Cl}^{-}\right]\).

Step by step solution

01

Identify the reaction conditions

Given that the solubility of \(Cl_{2}\) (g) in water is 6.4 g/L at 25°C, we are also given that \(K_c = 4.4 \times 10^{-4}\) for the reaction \(Cl_{2} (aq) + 2 H_{2}O (l) \rightarrow HOCl (aq) + H_{3}O^{+} (aq) + Cl^{-} (aq)\). It is important to note that because concentrations are given in g/L and need to be in M (mol/L), it is necessary to convert the solubility of \(Cl_{2}\) to molarity using the molar mass of chlorine. The molar mass of \(Cl_{2}\) is approximately 70.9 g/mol.

02

Determine initial and equilibrium concentrations

Thanks to the given solubility, we know \(Cl_{2}\) has an initial concentration of \( \frac{6.4 g/L}{70.9 g/mol} = 0.090 M \). Since at the start of the reaction, there are no products \(HOCl\), \(H_{3}O^{+}\) and \(Cl^{-}\) yet, their initial concentrations are zero. Letting x to be the change in concentration, at equilibrium we will have \(0.090 - x\) M of \(Cl_{2}\), and x M of \(HOCl\), \(H_{3}O^{+}\), \(Cl^{-}\) each.

03

Apply the equilibrium constant expression

The equilibrium constant expression for the reaction is \(K_c = \frac{[HOCl][H_{3}O^{+}][Cl^{-}]}{[Cl_{2}][H_{2}O]}\). Since the concentration of water is essentially treated as constant because water is in large excess (it's the solvent), and it's also a pure liquid, its concentration does not appear in the expression for equilibrium constant. Thus we have \(4.4 \times 10^{-4} = \frac{x^3}{0.090 - x}\).

04

Solve for the change in concentration

Solving this equation for x requires approximation because \(x\) is so small relative to 0.090 that the subtraction can be ignored. Thus, the equation becomes \(4.4 \times 10^{-4} = \frac{x^3}{0.090}\). Solving for x gives \(x = 6.3 \times 10^{-2}\) M.

05

Calculate concentrations of each species at equilibrium

Substituting \(x = 6.3 \times 10^{-2}\) back into the equilibrium expressions for each molecule gives \([Cl_{2}] = 0.090 - 6.3 \times 10^{-2} = 0.027 \) M, \([HOCl] = [H_{3}O^{+}] = [Cl^{-}] = 6.3 \times 10^{-2}\) M.

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