The solubility of \(\mathrm{Cl}_{2}\) in \(100 \mathrm{~g}\) of water at \(\mathrm{STP}\) is \(310 \mathrm{~cm}^{3}\). Assume that this quantity of \(\mathrm{Cl}_{2}\) is dissolved and equilibrated as follows: $$ \mathrm{Cl}_{2}(a q)+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{Cl}^{-}(a q)+\mathrm{HClO}(a q)+\mathrm{H}^{+}(a q) $$ (a) If the equilibrium constant for this reaction is \(4.7 \times 10^{-4}\), calculate the equilibrium concentration of \(\mathrm{HClO}\) formed. \((\mathbf{b})\) What is the \(\mathrm{pH}\) of the final solution?

Given that the solubility of Cl2 is 310 cm³ in 100 g of water (100 mL) at STP: 1 mol of any gas at STP = 22.4 L = 22400 cm³ Therefore, \( moles\, of\, Cl2 = \frac{310}{22400} = 0.013839 \, moles \)

The balanced equation is: \( Cl2(aq) + H2O \rightleftharpoons Cl^{-}(aq) + HClO(aq) + H^{+}(aq) \) Let the concentration of Cl2 at equilibrium be x moles, then: - [Cl2] = \(0.013839 - x\) - [Cl-] = [HClO] = [H+] = x Equilibrium constant, K = \(4.7 \times 10^{-4}\) Now, setting up the equilibrium expression: \(K = \frac{[Cl^{-}][HClO][H^{+}]}{[Cl2]} \) Since [Cl-] = [HClO] = [H+], the expression becomes: \( K = \frac{x^3}{0.013839-x} \) Replacing K with the given value and solving for x: \( 4.7 \times 10^{-4} = \frac{x^3}{0.013839-x} \)

To find the equilibrium concentration of HClO and H+ ions, we need to solve the equation for x. In this case, we can use a numerical method or an approximation. For simplicity and practical purposes, we'll use an approximation because K is considerably small: Assuming \(x \ll 0.013839\), \( 4.7 \times 10^{-4} \approx \frac{x^3}{0.013839} \) Now, solving for x: \( x^3 = 4.7 \times 10^{-4} \times 0.013839 \) \( x = \sqrt[3]{6.503 \times 10^{-6}} \) \( x = 1.868 \times 10^{-2} \, moles\) Since [HClO] = [H+] = x, the equilibrium concentration of HClO is \(1.868 \times 10^{-2} \, moles\, L^{-1}\) and the concentration of H+ ions is also \(1.868 \times 10^{-2} \, moles\, L^{-1}\). To find the pH of the resulting solution, use the relationship: \( pH = - \log[H^+] \) Thus, \( pH = - \log(1.868 \times 10^{-2}) \) \( pH \approx 1.73 \) Therefore, the equilibrium concentration of HClO is approximately \(1.868 \times 10^{-2} \, moles\, L^{-1}\) and the pH of the final solution is approximately 1.73.