Problem 44 - Chemistry The Science in Context

Question

A \(25.0 \mathrm{mL}\) sample of a \(0.100 M\) solution of aqueous trimethylamine is titrated with a \(0.125 M\) solution of \(\mathrm{HCl}\). What is the pH of the solution after \(10.0,20.0,\) and \(30.0 \mathrm{mL}\) of acid have been added?

Step-by-Step Solution

Answer

Using the titration method and pH calculations, determine the pH of a \(0.100 M\) aqueous trimethylamine solution after adding different volumes of \(0.125 M \mathrm{HCl}\) to the solution. After adding \(10.0 \mathrm{mL}\) of \(\mathrm{HCl}\), the pH of the solution is 9.80. After adding \(20.0 \mathrm{mL}\) of \(\mathrm{HCl}\), the pH of the solution is 9.80. After adding \(30.0 \mathrm{mL}\) of \(\mathrm{HCl}\), the pH of the solution is 1.64.

Step 1: Calculate the initial moles of trimethylamine

First, we need to determine the initial moles of trimethylamine in the \(25.0 \mathrm{mL}\) sample. We are given that the concentration of the solution is \(0.100 M\). To find the moles of trimethylamine, we can use the following formula: Moles of trimethylamine \(= \mathrm{volume (L)} \times \mathrm{molarity}\) Convert the volume of the solution from milliliters to liters: \(25.0 \mathrm{mL} = 0.025 \mathrm{L}\). Now, we can calculate the initial moles of trimethylamine: Moles of trimethylamine \(= 0.025 \mathrm{L} \times 0.100 \mathrm{M} = 0.00250 \mathrm{mol}\)

Step 2: Calculate the moles of HCl added at each specified volume

We are given that the concentration of the HCl solution is \(0.125 M\). To find the moles of HCl added at each specified volume, we can use the same formula as in Step 1: Moles of HCl \(= \mathrm{volume (L)} \times \mathrm{molarity}\) For each specified volume, convert the volume from milliliters to liters and calculate the moles of HCl: \(10.0 \mathrm{mL} = 0.01 \mathrm{L}\) Moles of HCl at \(10.0 \mathrm{mL} = 0.01 \mathrm{L} \times 0.125 \mathrm{M} = 0.00125 \mathrm{mol}\) \(20.0 \mathrm{mL} = 0.02 \mathrm{L}\) Moles of HCl at \(20.0 \mathrm{mL} = 0.02 \mathrm{L} \times 0.125 \mathrm{M} = 0.00250 \mathrm{mol}\) \(30.0 \mathrm{mL} = 0.03 \mathrm{L}\) Moles of HCl at \(30.0 \mathrm{mL} = 0.03 \mathrm{L} \times 0.125 \mathrm{M} = 0.00375 \mathrm{mol}\)

Step 3: Calculate the moles and concentrations of species in the solution after each addition

In this step, we will calculate the moles of trimethylamine and HCl after adding each specified volume of HCl to the trimethylamine solution. After each addition, we will determine the concentration of the species in solution. 1. After adding \(10.0 \mathrm{mL}\) of HCl: 0.00125 moles of HCl react with 0.00125 moles of trimethylamine, leaving 0.00125 moles of unreacted trimethylamine and forming 0.00125 moles of \(\mathrm{NH(CH_3)_3H^{+}}\). The total volume of the solution becomes 35.0 mL = 0.035 L. [Trimethylamine]: \(\frac{0.00125 \mathrm{mol}}{0.035 \mathrm{L}} = 0.0357 \mathrm{M}\) [\(\mathrm{NH(CH_3)_3H^{+}}\)]: \(\frac{0.00125 \mathrm{mol}}{0.035 \mathrm{L}}= 0.0357 \mathrm{M}\) 2. After adding \(20.0 \mathrm{mL}\) of HCl: 0.00250 moles of HCl completely react with 0.00250 moles of trimethylamine, leaving no unreacted trimethylamine and forming 0.00250 moles of \(\mathrm{NH(CH_3)_3H^{+}}\). The total volume of the solution becomes 45.0 mL = 0.045 L. [Trimethylamine]: 0 M [\(\mathrm{NH(CH_3)_3H^{+}}\)]: \(\frac{0.00250\mathrm{mol}}{0.045 \mathrm{L}}=0.0556 \mathrm{M}\) 3. After adding \(30.0 \mathrm{mL}\) of HCl: 0.00250 moles of trimethylamine react with 0.00250 moles of HCl. The remaining HCl is 0.00125 moles. There's no unreacted trimethylamine in the solution. The total volume of the solution becomes 55.0 mL = 0.055 L. [HCl]: \(\frac{0.00125 \mathrm{mol}}{0.055 \mathrm{L}} = 0.0227 \mathrm{M}\)

Step 4: Calculate the pH after each addition

Now we will calculate the pH of the solution after each addition. For the first and the second additions, we'll use the Henderson-Hasselbalch equation for the \(\mathrm{NH(CH_3)_3H^{+}}\)-trimethylamine buffer: $$ \mathrm{pH}=\mathrm{p}K_{a}+\log \frac{[\mathrm{Trimethylamine}]}{[\mathrm{NH(CH_3)_3H^{+}}]} $$ The pKa of trimethylamine is 9.80. 1. After adding \(10.0 \mathrm{mL}\) of HCl: $$\mathrm{pH} = 9.80 + \log \frac{0.0357}{0.0357} = 9.80$$ 2. After adding \(20.0 \mathrm{mL}\) of HCl: $$\mathrm{pH} = 9.80 + \log \frac{0}{0.0556}$$ Since the trimethylamine has been completely consumed, this formula isn't useful. However, since \(\mathrm{NH(CH_3)_3H^{+}}\) is the only species remaining, the pH is simply its pKa: $$\mathrm{pH} = 9.80$$ 3. After adding \(30.0 \mathrm{mL}\) of HCl: The solution now contains excess HCl. To find the pH, we can use the following formula: $$\mathrm{pH} = -\log [\mathrm{H^+}] = -\log [\mathrm{HCl}]$$ $$\mathrm{pH} = -\log 0.0227 = 1.64$$ So, the pH of the solution after adding \(10.0 \mathrm{mL}\), \(20.0 \mathrm{mL}\), and \(30.0 \mathrm{mL}\) of \(\mathrm{HCl}\) are 9.80, 9.80, and 1.64, respectively.

Key Concepts

Understanding Titration

Titration is a method in chemistry used to determine the concentration of a dissolved substance. It typically involves the gradual addition of a reactant solution (titrant) into a solution containing the substance being measured. This process continues until a reaction occurs at an equivalence point where the reactant has completely reacted with the substance in the solution. In our example, trimethylamine is titrated with hydrochloric acid (HCl).
The goal is to measure the amount of HCl needed for it to react completely with the trimethylamine. By knowing how many moles of HCl are needed, we can calculate concentrations and, subsequently, pH levels.

• The titrant here is \(0.125 \, M\) HCl, added in different volumes: 10 mL, 20 mL, and 30 mL.
• The end results are used to calculate the pH of the solution, giving insights into the behavior of the acid-base reaction.

By precisely measuring and recording the change in solution's pH as HCl is added, one can achieve an accurate understanding of the reaction dynamics and concentrations involved.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is an important tool in chemistry, especially for calculating the pH of buffer solutions. A buffer solution is one where a weak acid and its conjugate base or a weak base and its conjugate acid are present in comparable concentrations, which allows the solution to resist changes in pH upon the addition of small amounts of acid or base.
The equation can be expressed as:
\[ \text{pH} = \text{p}K_a + \log \left( \frac{[\text{Base}]}{[\text{Acid}]} \right) \] Here, \(\text{p}K_a\) is the negative logarithm of the acid dissociation constant, a measure of the acid's strength.

• In the exercise, after adding 10 mL of HCl, both trimethylamine and its conjugate acid, \(\mathrm{NH(CH_3)_3H^{+}}\), are present in equal concentrations, making \(\text{pH} = \text{p}K_a = 9.80\).
• For 20 mL, the equation cannot be directly used because there is no trimethylamine left, resulting in a solution dominated by the conjugate acid, keeping the pH at approximately its \(\text{p}K_a\).

This equation beautifully connects the chemical properties of substances with their practical measurements, simplifying pH calculation in buffer scenarios.

Basics of Acid-Base Chemistry

Acid-base chemistry focuses on the properties and reactions of acids and bases. Acids are substances that can donate protons (H⁺ ions), while bases are substances that can accept them. They react with each other in a neutralization reaction to form a salt and water.
Key to understanding this chemistry is the concept of pH, a measure on a scale ranging from 0 to 14 that expresses the acidity or basicity of a solution:

• A solution is acidic if its pH is less than 7.
• A neutral solution has a pH of 7.
• A basic solution has a pH greater than 7.

In titration, understanding the changes in pH as an acid or base is added enables precise calculations of substance concentrations. For example, trimethylamine reacts with HCl to form its corresponding salt, and the pH changes reflect the relative concentrations of these reactants and products. As HCl is added and trimethylamine is consumed, the nature of the solution shifts from basic to acidic, as seen after adding 30 mL, where excess HCl results in a very low pH indicating an acidic environment. Exploring these changes reveals the underlying dynamics of acid-base reactions and their equilibrium.