Consider the titration of \(30.0 \mathrm{~mL}\) of $0.050 \mathrm{M} \mathrm{NH}_{3}\( with \)0.025 M$ HCl. Calculate the pH after the following volumes of titrant have been added: $(\mathbf{a}) 0 \mathrm{~mL},(\mathbf{b}) 20.0 \mathrm{~mL},\( (c) 59.0 \)\mathrm{mL},(\mathbf{d}) 60.0 \mathrm{~mL},(\mathbf{e}) 61.0 \mathrm{~mL},(\mathbf{f}) 65.0 \mathrm{~mL} .$

First, we need to find the number of moles of NH₃ and HCl for each given volume. We can use the formula n = CV, where n is the number of moles, C is the concentration, and V is the volume. _initial moles of NH₃:_ n(NH₃) = \(0.050 M \times 30.0 mL = 1.5 \times 10^{-3} moles\) For volumes of HCl: a) 0 mL: n(HCl) = 0 moles b) 20.0 mL: n(HCl) = \(0.025 M \times 20.0 mL = 5 \times 10^{-4} moles\) c) 59.0 mL: n(HCl) = \(0.025 M \times 59.0 mL = 1.475 \times 10^{-3} moles\) d) 60.0 mL: n(HCl) = \(0.025 M \times 60.0 mL = 1.5 \times 10^{-3} moles\) e) 61.0 mL: n(HCl) = \(0.025 M \times 61.0 mL = 1.525 \times 10^{-3} moles\) f) 65.0 mL: n(HCl) = \(0.025 M \times 65.0 mL = 1.625 \times 10^{-3} moles\)

Now, we will find the moles of each species (NH₃, NH₄⁺, OH⁻, and Cl⁻) present in the solution after adding the given volume of titrant. \(a)\): no HCl NH₃ = 1.5 × 10^(-3) moles, NH₄⁺ = 0 moles, OH⁻ = 0 moles, Cl⁻ = 0 moles \(b)\): 20.0 mL HCl NH₃ = \(1.5 × 10^{-3} - 5 × 10^{-4}\) moles, NH₄⁺ = 5 × 10^(-4) moles, OH⁻ = 0 moles, Cl⁻ = 5 × 10^(-4) moles \(c)\): 59.0 mL HCl NH₃ = \(1.5 × 10^{-3} - 1.475 × 10^{-3}\) moles, NH₄⁺ = 1.475 × 10^(-3) moles, OH⁻ = 0 moles, Cl⁻ = 1.475 × 10^(-3) moles \(d)\): 60.0 mL HCl (equivalence point) NH₃ = 0 moles, NH₄⁺ = 1.5 × 10^(-3) moles, OH⁻ = 0 moles, Cl⁻ = 1.5 × 10^(-3) moles \(e)\): 61.0 mL HCl (post equivalence point) NH₃ = 0 moles, NH₄⁺ = \(1.5 × 10^{-3} + 0.025 × 10^{-3}\) moles, OH⁻ = 0 moles, Cl⁻ = 1.525 × 10^(-3) moles \(f)\): 65.0 mL HCl NH₃ = 0 moles, NH₄⁺ = \(1.5 × 10^{-3} + 0.125 × 10^{-3}\) moles, OH⁻ = 0 moles, Cl⁻ = 1.625 × 10^(-3) moles

To find the concentrations of each species, we need to divide the number of moles of each species by the total volume of the mixture at each given point. Final volume at each point: a) 30.0 mL b) 50.0 mL c) 89.0 mL d) 90.0 mL e) 91.0 mL f) 95.0 mL Using these final volumes, we can calculate the concentrations of each species at each point.

Using the concentrations of each species, we can calculate the H⁺ concentration using the dissociation constant formula (\(\$K_{a} = [H^{+}][NH_{3}] / [NH_{4}^{+}]\$) and subsequently calculate the pH using the pH formula (\(\$pH = -\log[H^{+})\]$. a) The H⁺ concentration comes from the dissociation of NH₃. After calculating H⁺ concentration, find pH using the pH formula. b) The H⁺ concentration comes from the reaction between NH₃ and NH₄⁺. Use the reaction equation to find H⁺ concentration and then find the pH using the pH formula. c) The H⁺ concentration comes from the reaction between NH₃ and NH₄⁺. Like in step b, find the pH using the pH formula. d) The H⁺ concentration comes from the dissociation of NH₄⁺. Use the dissociation constant formula, followed by the pH formula to find the pH. e) The H⁺ concentration comes from the excess HCl added to the solution. After finding the H⁺ concentration, use the pH formula to find the pH. f) Similar to step e, find the pH from the excess HCl and the pH formula. After calculating the pH at each titrant volume, you will have the corresponding pH values for each volume given in the exercise.