Problem 46 - Chemistry: The Central Science in SI Units, Global Edition

Question

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 \(\mathrm{mL},(\mathbf{d}) 60.0 \mathrm{~mL},(\mathbf{e}) 61.0 \mathrm{~mL},(\mathbf{f}) 65.0 \mathrm{~mL} .\)

Step-by-Step Solution

Answer

The pH values after adding the specified volumes of titrant are approximately: a) pH = 11.67 b) pH = 9.80 c) pH = 5.54 d) pH = 4.75 e) pH = 3.76 f) pH = 3.39

Step 1: Calculate moles of NH₃ and HCl for each volume

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) = \(0.025 M \times 65.0 mL = 1.625 \times 10^{-3} moles\)

Step 2: Calculating the number of moles of each species at each volume

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

Step 3: Calculating the concentrations of each species at each volume

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.

Step 4: Calculating pH at each volume

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.

Key Concepts

Calculating pH

The calculation of pH in a titration process is pivotal to understanding the progression of the reaction. pH, or "potential of hydrogen," measures the acidity or basicity of a solution. It is derived from the concentration of H⁺ ions in the mixture. To calculate the pH at different stages of a titration, we use the formula

• \( pH = -\log[H^+] \)

At the beginning of the titration (0 mL HCl added), the pH arises from the basicity of ammonia (\( NH_3 \)). As we proceed by adding HCl, an acid, it reacts with the ammonia, decreasing the pH. This ongoing neutralization captures the heart of acid-base chemistry. As we reach the equivalence point, all ammonia has reacted, making it crucial to compute the concentration of remaining species to derive pH.
Beyond the equivalence point, pH is determined by leftover HCl, reshaping the nature of the solution to more acidic.

Titration curve

A titration curve visually represents the pH change observed during a titration. It is essential for illustrating how a solution responds to the gradual addition of a titrant. When plotting a titration curve, the x-axis represents the volume of titrant added (e.g., HCl), while the y-axis measures the pH of the solution.
These curves provide critical insights:

• Identify the buffering regions where pH changes slowly
• Pinpoint the equivalence point where the amount of titrant perfectly neutralizes the initial solution
• Highlight the steep regions signaling rapid pH shifts

For the titration of ammonia with hydrochloric acid, the curve starts at a relatively high pH, typical of bases like ammonia, and descends as HCl is introduced. The downward progression continues until reaching the equivalence point, beyond which the solution becomes acidic as excess HCl dominates.

Acid-base reaction

Acid-base reactions are cornerstone processes in chemistry where acid and base interact to form water and salt. In this exercise, we witness the reaction between ammonia (\( NH_3 \)) and hydrochloric acid (\( HCl \)).
Ammonia acts as a weak base and accepts protons from hydrochloric acid, a strong acid, resulting in the production of ammonium ions (\( NH_4^+ \)) and chloride ions (\( Cl^- \)). This specific reaction is represented by the following equation:

• \( NH_3 (aq) + HCl (aq) \rightarrow NH_4^+ (aq) + Cl^- (aq) \)

Understanding this interaction is fundamental to comprehending how titration measures the point of neutralization. As we progress by adding HCl to the ammonia solution, the pH progressively decreases until reaching neutrality at the equivalence point, where all ammonia is transformed.

Mole calculation

Mole calculations are a crucial step in titration problems, equipping you with the numerical understanding necessary for stoichiometric balance. The core idea here is to determine how many moles of each reactant and product are present or created during the titration process.
Use the formula

• \( n = C \times V \)

where \( n \) is the number of moles, \( C \) is the concentration, and \( V \) is the volume in liters. By applying this formula:

• Initial moles of \( NH_3 \) at 0 mL of HCl are designed, knowing its concentration and starting volume.
• As HCl is added, calculate its moles at each step to track the reaction progress.

These calculations enable us to determine how much \( NH_3 \) and \( NH_4^+ \) are present after each addition of HCl, which subsequently influences the resulting pH and shapes the titration curve. Mole calculations are not only foundational for chemistry beans but also for a clear visualization of the chemical reaction throughout the titration process.