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

To determine the moles of \(NH_3\), we'll use the Ideal Gas Law, which states: \(PV=nRT\), where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin. First, convert the given values to the appropriate units: - Volume (\(V\)): \(7.5 L\) - Temperature (\(T\)): \(22^{\circ}C + 273.15 = 295.15K\) - Pressure (\(P\)): \(\frac{735 torr}{760} = 0.967atm\) Next, we'll use the value for R in atm, which is \(0.0821 \frac{L \cdot atm}{mol \cdot K}\): \(0.967atm \cdot 7.5L = n \cdot 0.0821 \frac{L \cdot atm}{mol \cdot K} \cdot 295.15K\) Now, we solve for n: \(n = \frac{0.967atm \cdot 7.5L}{0.0821 \frac{L \cdot atm}{mol \cdot K} \cdot 295.15K} = 0.297 mol\)

We are given the concentration and volume of the HCl solution: Concentration of HCl: \(0.40M\) Volume of HCl solution: \(0.50L\) Now we can find the moles of HCl using the formula: Moles of HCl = Concentration × Volume Moles of HCl = \(0.40M \times 0.50L = 0.2 mol\)

\(^NH_3\) and HCl react in a 1:1 molar ratio: \(NH_3 + HCl \rightarrow NH_4Cl\) Since we have more moles of \(NH_3\) than HCl, we will have some leftover \(NH_3\) after the reaction. To find out how much, subtract the moles of HCl from the moles of \(NH_3\): Remaining moles of \(NH_3 = 0.297 \text{mol} - 0.2 \text{mol} = 0.097 \text{mol}\)

We are told that the volume of the solution remains constant at \(0.50L\). To find the concentration of the leftover \(NH_3\), we use the formula: Concentration = Moles ÷ Volume Concentration of \(NH_3 = \frac{0.097 \text{mol}}{0.50L} = 0.194M\)

Since \(NH_3\) is a weak base, it will react with the water in the solution, forming \(OH^-\). To find the concentration of \(OH^-\), we can use the formula with the base ionization constant (\(K_b\)) of \(NH_3\), which is approximately \(1.8 \times 10^{-5}\): \(K_b = [NH_4^+][OH^-] / [NH_3]\) Assuming that the change in concentrations due to the reaction is negligible, we have: \(1.8 \times 10^{-5} = [x][x] / [0.194]\) Solving for x, we get: \([OH^-] = x = 1.83 \times 10^{-3}\) Now, we'll use the relationship between \(OH^-\) and \(H^+\) concentrations to determine the \(H^+\) concentration: \(K_w = [H^+][OH^-]\), where \(K_w = 1 \times 10^{-14}\) \([H^+] = \frac{1 \times 10^{-14}}{1.83 \times 10^{-3}} = 5.46 \times 10^{-12}\) Finally, we can calculate the pH using the formula: pH = \(-\log[H^+]\) pH = \(-\log(5.46 \times 10^{-12}) = 11.26\) The pH of the resulting solution is approximately 11.26.