The solubility of \(\mathrm{Cl}_{2}\) in 100 \(\mathrm{g}\) of water at 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 HClO formed. (b) What is the pH of the final solution?

To find out the moles of Cl2 in water, we first need to convert the volume of Cl2 from cm³ to L and then use the equation: moles of Cl2 = PV/RT where P = pressure (1 atm) V = volume (Convert the volume of Cl2 from cm³ to L: 310 cm³ * (1 L/1000 cm³) = 0.31 L) R = gas constant (0.0821 L atm / K mol) T = temperature (273 K) Moles of Cl2 = (1 atm * 0.31 L) / (0.0821 L atm / K mol * 273 K)

Based on the stoichiometry of the balanced equation, the moles of Cl2 is equal to the moles of HClO formed. Total mass of solvent = 100 g water Molar mass of water = 18 g/mol Moles of solvent (water) = 100 g / 18 g/mol = 5.56 mol Now we calculate the initial concentration of Cl2: Initial concentration = Moles of Cl2 / Moles of solvent (5.56 mol) Using the calculated initial concentration of Cl2 and equilibrium constant (4.7 x 10^-4) given in the problem, we can now find out the equilibrium concentration of HClO formed. The reaction stoichiometry suggests the following relationships at equilibrium: [Cl2] = [Cl2]0 - [HClO] [Cl-] = [HClO] [HClO] = [H+] [Cl2] = Initial (Cl2) concentration - [HClO] Plugging these relationships into the equilibrium constant expression, we get: K = ([Cl-][HClO][H+])/([Cl2][H2O]) Since the concentration of H2O is large and relatively constant, we can simplify this expression to: K' = ([HClO]^3)/([Cl2]0 - [HClO]) Plug in K and initial concentration found earlier, then solve the cubic equation for [HClO].

Use the equilibrium concentration of H+ ([H+]) found in step 2 to calculate the pH of the final solution using the pH formula: pH = -log([H+]) By calculating the pH, we get the final answer to this problem.