At \(25^{\circ} \mathrm{C}\) and \(1.00\) atm pressure, one liter of ammonia is bubbled into \(725 \mathrm{~mL}\) of water. Assume that all the ammonia dissolves and the volume of the solution is the volume of the water. A \(50.0-\mathrm{mL}\) portion of the prepared solution is titrated with \(0.2193 \mathrm{M} \mathrm{HNO}_{3} .\) Calculate the \(\mathrm{pH}\) of the solution (a) before titration. (b) halfway to the equivalence point. (c) at the equivalence point.

Use the Ideal Gas Law to find the moles of ammonia (\(n_{NH_3}\)), given the initial volume, pressure and temperature of ammonia gas: \(PV = nRT\) where: P = 1.00 atm (pressure) V = 1 L (volume of ammonia gas) R = 0.0821 L atm/(mol K) (ideal gas constant) T = 25^{\circ}C + 273.15 = 298.15 K (temperature in Kelvin) Solve for \(n_{NH_3}\): \(n_{NH_3} = \frac{PV}{RT} = \frac{1.00 \ \text{atm} \cdot 1 \ \text{L}}{0.0821 \ \text{L atm} / (\text{mol K}) \cdot 298.15 \ \text{K}} \approx 0.0407 \ \text{moles}\)

Calculate the concentration of ammonia (\([\mathrm{NH_3}]\)) in the solution after bubbling into water. We will assume that the volume of water (725 mL) is equal to the volume of the solution: \([\mathrm{NH_3}] = \frac{n_\mathrm{NH_3}}{V_\mathrm{solution}} = \frac{0.0407 \ \text{moles}}{725 \ \text{mL}} \approx 5.62 \times 10^{-2} \ \mathrm{M}\)

Ammonia is a weak base (refer to a table of Kb values or use given information), so we can set up an ICE (Initial, Change, Equilibrium) table to determine the concentration of hydroxide ions, \(\mathrm{OH^-}\), at equilibrium: $\begin{array}{c|cccc} & \mathrm{NH_3} & + & \mathrm{H_2O} & \rightleftharpoons & \mathrm{NH_4^+} & + & \mathrm{OH^-} \\ \hline \text{Initial} & 5.62\times10^{-2} \ \mathrm{M} & & & & 0 & & 0 \\ \text{Change} & -x & & & & +x & & +x \\ \text{Equilibrium} & 5.62\times10^{-2} \ \mathrm{M} -x & & & & x & & x \\ \end{array}$ We have: Kb (for ammonia) = \(1.8 \times 10^{-5}\) Hence, we can write the equilibrium expression: \(K_\mathrm{b} = \frac{[\mathrm{NH_4^+}][\mathrm{OH^-}]}{[\mathrm{NH_3}]}\)

(a) Before titration: Using the weak base equilibrium, we can find the concentration of \(\mathrm{OH^-}\) which will allow us to calculate the pH of the solution: \(1.8 \times 10^{-5} = \frac{x^2}{5.62 \times 10^{-2}-x}\) Since Kb is very small, we can assume \(x\ll 5.62\times 10^{-2}\): \(1.8\times 10^{-5} \approx \frac{x^2}{5.62 \times 10^{-2}}\) Solving for x (which represents the concentrations of \(\mathrm{NH_4^+}\) and \(\mathrm{OH^-}\)): \(x \approx \mathrm{OH^-} \cong \sqrt{(1.8 \times 10^{-5})(5.62 \times 10^{-2})} \approx 1.8\times10^{-3} \ \mathrm{M}\) Now, calculate the \(\mathrm{pOH}\) and \(\mathrm{pH}\): \(\mathrm{pOH} = -\log_{10} [\mathrm{OH^-}] \approx -\log_{10} (1.8\times10^{-3}) \approx 2.74\) \(\mathrm{pH} = 14 - \mathrm{pOH} \approx 14 - 2.74 \approx 11.26\) The pH of the solution before titration is approximately 11.26. (b) Halfway to the equivalence point: At halfway to the equivalence point, the ratio \([\mathrm{NH_4^{+}}]/[\mathrm{NH_3}] = 1\). Using the given Kb expression: \(K_\mathrm{b} = \frac{[\mathrm{OH^-}]}{[\mathrm{NH_3}]} = 1.8\times10^{-5}\) Therefore: \([\mathrm{OH^-}] = 1.8 \times 10^{-5} \ \mathrm{M}\) Now, calculate the pOH and pH: \(\mathrm{pOH} = -\log_{10} [\mathrm{OH^-}] \approx -\log_{10} (1.8\times10^{-5}) \approx 4.74\) \(\mathrm{pH} = 14 - \mathrm{pOH} \approx 14 - 4.74 \approx 9.26\) The pH of the solution halfway to the equivalence point is approximately 9.26. (c) At the equivalence point: At the equivalence point, all the ammonia has reacted with the acid. We need to find how much acid is required. Since the initial concentration of ammonia is known, we can calculate the moles of ammonia: \(n_{NH_3} = (5.62\times10^{-2}\ \mathrm{M})\times(0.050\ \mathrm{L}) = 2.81\times10^{-3}\ \mathrm{moles}\) We also know: \(n_{HNO_3} = n_{NH_3}\) Now, we can find the volume of \(\mathrm{HNO_3}\) required: \(V_{HNO_3} = \frac{n_{HNO_3}}{C_{HNO_3}} = \frac{2.81\times 10^{-3}\ \mathrm{moles}}{0.2193\ \mathrm{M}} \approx 0.0128\ \mathrm{L}\) Since there is no \(\mathrm{OH^-}\) left at the equivalence point, the remaining acidic species will determine the pH. In this case, the remaining acidic species is \(\mathrm{NH_4^+}\) (ammonium ion). The concentration of \(\mathrm{NH_4^+}\) at the equivalence point can be determined as: \([\mathrm{NH_4^+}] = \frac{n_{NH_4^+}}{V_{total}} = \frac{2.81\times 10^{-3}\ \mathrm{moles}}{50.0\ \mathrm{mL} + 12.8\ \mathrm{mL}} \approx 4.26 \times 10^{-2} \ \mathrm{M}\) Now, we need to calculate the concentration of \(\mathrm{H^+}\) from the dissociation of \(\mathrm{NH_4^+}\). We are given the Ka value for \(\mathrm{NH_4^+}\), which is 5.6 × 10^(-10). Using the Ka expression: \(K_\mathrm{a} = \frac{[\mathrm{H^+}][\mathrm{NH_3}]}{[\mathrm{NH_4^+}]}\) Since the dissociation of \(\mathrm{NH_4^+}\) is small, we can assume that: \([\mathrm{H^+}] \approx [\mathrm{NH_3}]\) Hence, \(5.6 \times 10^{-10} = \frac{x^2}{4.26 \times 10^{-2} - x}\) Since Ka is very small, we can assume \(x\ll 4.26\times 10^{-2}\): \(5.6 \times 10^{-10} \approx \frac{x^2}{4.26 \times 10^{-2}}\) Solving for x: \([\mathrm{H^+}] \approx x \approx \sqrt{(5.6 \times 10^{-10})(4.26 \times 10^{-2})} \approx 3.6\times 10^{-6} \ \mathrm{M}\) Finally, calculate the pH: \(\mathrm{pH} = - \log_{10} [\mathrm{H^+}] \approx -\log_{10} (3.6\times10^{-6}) \approx 5.44\) The pH of the solution at the equivalence point is approximately 5.44.