Question

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

Short answer

The pH values at the specified stages of the titration are as follows: (a) 0 mL: \( \mathrm{pH = 11.1} \) (b) 20 mL: \( \mathrm{pH \approx 9.3} \) (c) 59 mL: \( \mathrm{pH \approx 4.70} \) (d) 60 mL: \( \mathrm{pH \approx 4.2} \) (e) 61 mL: \( \mathrm{pH \approx 3.82} \) (f) 65 mL: \( \mathrm{pH \approx 3.42} \)

Step-by-step solution

Calculate moles of NH3

First, we need to calculate the initial moles of NH3 before the titration starts. Initial moles of NH3 = Molarity × Volume = 0.050 mol/L * 0.030 L = 0.0015 mol

Calculate the new concentrations after the titrant has been added at the given points

For every volume of titrant, we need to calculate the new moles and concentrations of NH3, HCl, and NH4Cl in the solution. For this, we will use the following equations: 1. Moles of HCl = Molarity × Volume (in Liters) added. 2. Moles of NH3 = Initial moles - Moles of HCl reacted. 3. Moles of NH4Cl = Moles of HCl reacted. 4. New concentration = Moles / (30 mL + Volume of HCl added).

Calculate the pH at given points of the titration

Now, we will need to calculate the pH at the given points of titration as follows: (a) 0 mL => No HCl has been added, so pH can be directly calculated using the Kb of NH3. (b) 20 mL => At this point, NH3, HCl, and NH4Cl have reacted, and we have a buffer system. We can use the Henderson-Hasselbalch equation to calculate the pH. (c) 59 mL => Here, the moles of HCl and NH3 are nearly equal, so pH can be calculated as the pKa of NH4+. (d) 60 mL => At this point, the moles of HCl slightly exceed the moles of NH3, so we have to calculate the H+ concentration using the \(\mathrm{HCl}\) that is in excess. (e) 61 mL => HCl is in excess, so we can use this to find the pH. (f) 65 mL => Similar to (e), excess HCl can be used to find the pH. By calculating the moles of each species in the solution and using the appropriate equations, we can find the pH at the specified stages of titration.

Key concepts

Acid-Base Titration

The process of acid-base titration involves gradually adding a solution of known concentration (titrant) to a solution of unknown concentration (analyte) to determine the analyte's concentration. This procedure is widely used in chemistry to analytically determine the molarity of an acid or base in a sample. Typically, the point at which neutralization occurs is identified using an indicator or a pH meter. In the case of titrating ammonia with hydrochloric acid, the neutralization reaction results in the formation of ammonium chloride, a salt, and water.

During titration, different phases are observed: initially, the solution is primarily composed of the weak base (in this case, NH₃), then it enters a buffer region as the acid is added, and finally, it becomes acidic once the base is completely neutralized.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a formula that relates the pH of a buffer solution to the pKa value of the acid and the concentrations of the acid and its conjugate base. It's given by the formula: \[ \text{pH} = \text{pKa} + \log\left(\frac{[\text{Base}]}{[\text{Acid}]}\right) \] This equation is incredibly valuable when dealing with buffer solutions during a titration, as it allows for the calculation of the pH when the solution consists of a weak acid or base and its salt. By using this equation, students can understand how the pH of the solution changes when the ratio of conjugate base to acid changes, as seen when NH₃ is titrated with HCl.

To apply the Henderson-Hasselbalch equation, it is essential to know the pKa of the acid, which in the case of the ammonium ion (NH₄⁺), is the conjugate acid of NH₃. In scenarios where the concentrations of the acid and conjugate base are equal, the pH equals the pKa.

Buffer Solution

A buffer solution resists changes in pH when small amounts of acid or base are added. This is due to the presence of a weak acid and its conjugate base, or a weak base and its conjugate acid. In the context of the given titration problem, when NH₃ (a weak base) reacts with HCl (a strong acid), the resultant NH₄⁺ acts as the conjugate acid forming a buffer solution at certain points during the titration.

Buffer solutions play a crucial role during the equivalence point in titrations because the pH changes only gradually. This stability is essential in many biological and chemical applications where maintaining a consistent pH is critical.

Molarity Calculations

Molarity represents the concentration of a solution and is defined as the number of moles of solute per liter of solution. It is expressed in units of moles per liter (mol/L). To calculate molarity, one can use the equation: \[ \text{Molarity} = \frac{\text{moles of solute}}{\text{liters of solution}} \] In the exercise, students are first asked to calculate the molarity of the NH₃ and HCl solutions at different points during the titration. Knowing the molarity of the reacting solutions is vital for predicting the amount of product formed and for determining the pH at various stages of the titration process.

pH Determination

The pH of a solution is a measure of its acidity or basicity, which is a key concept in acid-base titrations. The pH can be calculated by taking the negative logarithm of the hydrogen ion concentration in the solution: \[ \text{pH} = -\log[H^{+}] \] Different stages of the titration impact the pH in unique ways. Initially, the pH can be determined based on the molarity and the intrinsic properties of the weak base, NH₃. During the buffer stage, the pH is calculated using the Henderson-Hasselbalch equation considering the amounts of NH₃ and NH₄⁺ in the solution. After the equivalence point, when HCl is in excess, the pH can be found directly from the concentration of the excess H⁺ ions resulting from the dissociation of the strong acid, HCl.