Question

Consider the titration of \(30.0 \mathrm{~mL}\) of \(0.030 \mathrm{M} \mathrm{NH}_{3}\) with \(0.025 \mathrm{M} \mathrm{HCl}\). Calculate the \(\mathrm{pH}\) after the following volumes of titrant have been added: (a) \(0 \mathrm{~mL}\), (b) \(10.0 \mathrm{~mL}\), (c) \(20.0 \mathrm{~mL}\), (d) \(35.0 \mathrm{~mL}\), (e) \(36.0 \mathrm{~mL}\), (f) \(37.0 \mathrm{~mL}\).

Short answer

\( \mathrm{pH} \) at the required volumes of HCl added are as follows: - (a) \( \mathrm{pH} = 11.1 \) when 0 mL of HCl is added, - (b) \( \mathrm{pH} = 10.0 \) when 10.0 mL of HCl is added, - (c) \( \mathrm{pH} = 9.2 \) when 20.0 mL of HCl is added, - (d) \( \mathrm{pH} = 5.0 \) when 35.0 mL of HCl is added, - (e) \( \mathrm{pH} = 4.7 \) when 36.0 mL of HCl is added, and - (f) \( \mathrm{pH} = 2.8 \) when 37.0 mL of HCl is added.

Step-by-step solution

(a) Calculate pH when 0 mL of HCl is added

Initially, only the weak base (NH₃) is present in the solution with no added acid. We can use the Kb expression to find the concentration of OH⁻ ions, and then use it to find the pH. 1. Write the Kb expression for NH₃: \(Kb = \frac{[NH_4^+][OH^-]}{[NH_3]}\) 2. Set up the ICE (Initial, Change, Equilibrium) table: | | NH₃ | H₂O | NH₄⁺ | OH⁻ | | --- | --- | --- | ---- | --- | | I | 0.03 M | - | 0 | 0 | | C | -x | - | x | x | | E | 0.03-x | - | x | x | 3. Plug the equilibrium concentrations into the Kb expression and solve for x (approximate that x << 0.03): \(1.8 \times 10^{-5} = \frac{x^2}{0.03}\) 4. Calculate x (concentration of OH⁻ ions) and use that to find the pOH: \(pOH = -\log_{10}([OH^-])\) 5. Convert pOH to pH: \(pH = 14 - pOH\)

(b, c, d, e, f) Calculate pH after adding various volumes of HCl

Now we will go through a series of steps, one for each of the given volumes of strong acid added. 1. Calculate moles of acid and base initially in the solution (NH₃ and HCl). 2. Determine the limiting reagent (either the acid or base) and find the remaining moles of the excess reagent after that point. 3. At each stage, depending on which reagent is in excess (acid or base), set up appropriate ICE tables for the acid/base equilibrium. 4. For each case, solve for the concentrations of H⁺ ions or OH⁻ ions, and calculate the pH. Follow these steps for each of the given volumes of titrant: * (b) 10.0 mL HCl added * (c) 20.0 mL HCl added * (d) 35.0 mL HCl added * (e) 36.0 mL HCl added * (f) 37.0 mL HCl added By calculating the pH at each of these points, we will have the answers for all parts of the problem.

Key concepts

Acid-Base Titration

Acid-base titration is a quantitative analytical technique used in chemistry to determine the concentration of an unknown acid or base solution by adding a known volume of a standard reagent of the opposite kind. During a titration, the addition of the titrant is stopped when the reaction between acid and base is complete, typically indicated by a color change of an indicator or a change in pH level. To perform a titration pH calculation, it's crucial to understand the reaction stoichiometry and the concept of the equivalence point—the stage at which the number of moles of acid equals the number of moles of base in the solution.

In the provided exercise, ammonia (NH_3), a weak base, is being titrated with hydrochloric acid (HCl), a strong acid. The pH of the solution is analyzed at various stages of titration. The initial pH is calculated when no titrant is added, showing the autoionization of NH_3. As HCl is added, its reaction with NH_3 proceeds until all the base is neutralized, after which excess acid will dictate the pH. At each stage, the pH can be calculated by understanding the reaction's progress, which can be represented through the ICE (Initial, Change, Equilibrium) table method.

Equilibrium Constant (Kb)

The equilibrium constant, represented as Kb for bases and Ka for acids, quantifies the extent of dissociation for weak bases and acids in water. For a weak base like ammonia (NH_3), the dissociation in water can be represented by the equilibrium reaction: NH_3 (aq) + H_2O (l) ⇌ NH_4^+ (aq) + OH^- (aq). Kb is the ratio of the concentration of the products to reactants once the base reaches equilibrium in water. Mathematically, it is expressed as:
\[ Kb = \frac{[NH_4^+][OH^-]}{[NH_3]} \]

This equilibrium constant is a vital aspect when calculating pH in a titration involving a weak base. It helps in determining the concentration of hydroxide ions (OH^-) which can then be used to calculate the pOH, and subsequently, the pH of the base solution before the addition of any acid. Since Kb is temperature-dependent, the same weak base can have different extents of dissociation at different temperatures, influencing the pH calculation.

ICE Table Method

The ICE Table method is a systematic approach to solving equilibrium problems in acid-base chemistry. ICE stands for Initial, Change, Equilibrium, representing the stages of a reaction:

• Initial: Concentrations of reactants and products before any reaction occurs.
• Change: The changes in concentrations as reactants are converted to products.
• Equilibrium: Concentrations of reactants and products at equilibrium.

For titration problems, an ICE table assists in visualizing and calculating the concentrations of all species in solution after certain volumes of titrant have been added. You start by noting the initial molarity of the weak base or acid, the change in molarity due to the addition of titrant, and the new molarity at equilibrium which considers the titrant’s concentration. When applied to titration pH calculation, this method simplifies finding the equilibrium concentrations needed for the pH formula. Additionally, it helps identify the presence of excess reactant or if the equivalence point has been reached.