A sample of \(7.5 \mathrm{~L}\) of \(\mathrm{NH}_{3}\) gas at \(22{ }^{\circ} \mathrm{C}\) and 735 torr is bubbled into a 0.50 - \(\mathrm{L}\) solution of \(0.40 \mathrm{M} \mathrm{HCl}\). Assuming that all the \(\mathrm{NH}_{3}\) dissolves and that the volume of the solution remains \(0.50 \mathrm{~L},\) calculate the \(\mathrm{pH}\) of the resulting solution.

First, let's use the Ideal Gas Law (PV=nRT) to find the moles of ammonia. We are given the volume (V), temperature (T), and pressure (P) of the ammonia gas. R (gas constant) = 62.36 L·Torr/mol·K (use this value since the pressure is given in Torr) Rearrange the Ideal Gas Law to solve for n (moles of ammonia): n = PV / RT n = (735 Torr × 7.5 L) / (62.36 L·Torr/mol·K × (22 °C + 273.15)) n = 0.30093 moles of NH3

Now let's find the moles of HCl in the solution. We are given the volume and molarity of the HCl solution. moles of HCl = volume × molarity moles of HCl = 0.50 L × 0.40 mol/L moles of HCl = 0.20 moles of HCl

The reaction between ammonia and hydrochloric acid produces ammonium chloride (NH4Cl). Let's write the balanced chemical equation for the reaction: NH3 + HCl -> NH4+ + Cl-

Now let's perform stoichiometric calculations to find the amounts of NH4+ ions and leftover HCl in the solution after the reaction. Since ammonia is the limiting reactant, we will calculate the moles of NH4+ produced and subtract the moles of ammonia that reacted from the moles of hydrochloric acid to find the leftover moles of HCl. moles of NH4+ = 0.30093 moles (as it is 1:1 stoichiometry) moles of leftover HCl = moles of initial HCl - moles of reacted NH3 moles of leftover HCl = 0.20 moles - 0.30093 moles moles of leftover HCl = - 0.10093 moles (Note: Since we have a negative value, this means that there is no leftover HCl and we have 0 leftover moles of HCl)

Now that we have the moles of NH4+ ions and leftover HCl in the solution, let's calculate their concentrations. Since the volume of the solution remains 0.50 L, we will divide the moles by 0.50 L to find the concentration of the ions. concentration of NH4+ = moles of NH4+ / volume of solution concentration of NH4+ = (0.30093 moles) / 0.50 L concentration of NH4+ = 0.60186 M

Now that we have the concentration of NH4+ ions, we can calculate the pH of the solution using the following equations: \(K_a = \frac{[H^+][\mathrm{A}^-]}{ [\mathrm{HA}] }\) \(pK_w = pK_a + pK_b\) \(K_a\) for NH4+ is \(5.6\times{10^{-10}}\), given that: \(K_b\) for NH3 is \(1.8\times{10^{-5}}\) \(10^{-14} = pK_a + pK_b\) \(10^{-14} = K_a + 1.8\times 10^{-5}\) \(K_a = 5.6\times{10^{-10}} = \frac{[H^+][\mathrm{NH_3}]}{ [\mathrm{NH_4^+}] }\) Now we can solve for the concentration of H+ ions in the solution: \([H^+] = \frac{K_a [\mathrm{NH_4^+}]}{ [\mathrm{NH_3}] }\) As there is no HCl left, all NH3 reacted, thus, the remaining concentration of NH3 is 0: \([H^+] = \frac{(5.6\times{10^{-10}}) (0.60186 M)}{ 0 }\) Since the division by zero yields an undefined result, we can assume the resulting solution is acidic with a pH less than 7, as there is no remaining NH3 to act as a base competing with H+ ions or any other pH buffering component in the solution.