Problem 45 - Chemistry: Principles and Reactions

Question

A \(20.00\) -mL sample of \(0.220 \mathrm{M}\) triethylamine, (CH \(\left._{3} \mathrm{CH}_{2}\right)_{3} \mathrm{~N}\), is titrated with \(0.544 \mathrm{M} \mathrm{HCl}\). $$\left(\mathrm{Kb}\left(\mathrm{CH}_{3} \mathrm{CH}_{2}\right)_{3} \mathrm{~N}=5.2 \times 10^{-4}\right)$$ (a) Write a balanced net ionic equation for the titration. (b) How many \(\mathrm{mL}\) of \(\mathrm{HCl}\) are required to reach the equivalence point? (c) Calculate \(\left[\left(\mathrm{CH}_{3} \mathrm{CH}_{2}\right)_{3} \mathrm{~N}\right],\left[\left(\mathrm{CH}_{3} \mathrm{CH}_{2}\right)_{3} \mathrm{NH}^{+}\right],\left[\mathrm{H}^{+}\right]\) and \(\left[\mathrm{Cl}^{-}\right]\) at the equivalence point. (Assume that volumes are additive.) (d) What is the \(\mathrm{pH}\) at the equivalence point?

Step-by-Step Solution

Answer

At the equivalence point, the pH is approximately 5.86.

Step 1: (a) Balanced net ionic equation

The balanced net ionic equation for the titration is the reaction between the base (triethylamine) and the acid (HCl): $$(CH_3CH_2)_3N + H^+ \rightarrow (CH_3CH_2)_3NH^+$$

Step 2: (b) Volume of HCl required to reach equivalence point

To determine the volume of HCl needed to reach the equivalence point, we use the following relationship: $$ \textrm{moles of base} = \textrm{moles of acid} $$ Using the given information, we can find the moles of base: $$ \textrm{moles of base} = (20.0 \textrm{ mL})\left(\frac{0.220 \textrm{ mol}}{1 \textrm{ L}}\right) = 0.0044 \textrm{ mol} $$ Now we can find the volume of acid: $$ \textrm{volume of HCl} = \frac{\textrm{moles of base}}{\textrm{concentration of HCl}} = \frac{0.0044 \textrm{ mol}}{0.544 \textrm{ M}} = 0.00809 \textrm{ L} = 8.09 \textrm{ mL} $$

Step 3: (c) Concentrations of various species at equivalence point

At the equivalence point, the moles of the weak base and strong acid are equal, so the number of moles of triethylammonium ion, \((CH_3CH_2)_3NH^+\), formed is equal to that of the moles of triethylamine initially present: $$ [ (CH_3CH_2)_3NH^+ ] = \frac{0.0044 \textrm{ mol}}{20.0 \textrm{ mL} + 8.09 \textrm{ mL}} = \frac{0.0044 \textrm{ mol}}{28.09 \textrm{ mL} \times \frac{1 \textrm{ L}}{1000 \textrm{ mL}}} = 0.156 \textrm{ M} $$ At this point, there is no more free triethylamine left, so: $$ [ (CH_3CH_2)_3N ] = 0 $$ Also, the concentration of Cl\(^-\) ions is equal to the concentration of triethylammonium ions: $$ [\mathrm{Cl}^-] = 0.156 \textrm{ M} $$ To find the concentration of H\(^+\) ions, we use the expression for Kb and the relationship between Kb and Ka: $$ \textrm{Kb} = \frac{[\textrm{BH}^+][\textrm{OH}^-]}{[\textrm{B}]} $$ Since triethylamine is a weak base and its Kb is given, we can find its Ka: $$ \textrm{Kw} = 1 \times 10^{-14} = \textrm{Ka} \times \textrm{Kb} $$ $$ \textrm{Ka} = \frac{1 \times 10^{-14}}{5.2 \times 10^{-4}} \approx 1.92 \times 10^{-11} $$ Now, using Ka, we can find [H\(^+\)]: $$ \textrm{Ka} = \frac{[\textrm{H}^+][\textrm{A}^-]}{[\textrm{HA}]} $$ $$ 1.92 \times 10^{-11} = \frac{[\textrm{H}^+](0.156 \textrm{ M})}{[\textrm{H}^+]} $$ $$ [\mathrm{H}^{+}] = \sqrt{1.92 \times 10^{-11}} \approx 1.39 \times 10^{-6} \textrm{ M} $$

Step 4: (d) pH at the equivalence point

Now that we have the concentration of H\(^+\) ions, we can find the pH at the equivalence point using the pH formula: $$ \textrm{pH} = -\log [\textrm{H}^+] = -\log (1.39 \times 10^{-6}) = 5.86 $$ At the equivalence point, the pH is approximately 5.86.

Key Concepts

Net Ionic Equation

The net ionic equation is a simplified version of a chemical equation that shows only the species that actually change during the reaction. It is an essential tool for understanding the chemistry behind acid-base titrations. During a titration, a base such as triethylamine \( (CH_3CH_2)_3N \) reacts with an acid like hydrochloric acid \( HCl \) to form a salt. When \( HCl \) is dissolved in water, it dissociates completely into \( H^+ \) and \( Cl^- \) ions. Since \( HCl \) is a strong acid, the reaction proceeds to completion.

The balanced net ionic equation for the given titration would involve the triethylamine (a weak base) accepting a proton \( H^+ \) from the hydrochloric acid to form triethylammonium ion \( (CH_3CH_2)_3NH^+ \) as shown in the solution. By focusing on the net ionic equation, students can better understand the actual change occurring during the reaction without getting distracted by spectator ions like \( Cl^- \) which do not participate in reaction.

Acid-Base Titration

An acid-base titration is a process used to determine the concentration of a solution by reacting it with a solution of known concentration. The equivalence point of a titration is the point at which the amount of acid equals the amount of base in the solution. This concept is central to understanding titrations because it is the key to determining the concentration of the unknown solution.

In the context of the exercise, the equivalence point is reached when the moles of triethylamine are equal to the moles of \( HCl \) added. At this point, all of the triethylamine has been converted to triethylammonium ions, and the pH can be calculated. To improve understanding, students should visualize the titration curve and appreciate that at the equivalence point in the titration of a weak base with a strong acid, the solution contains the conjugate acid of the weak base, and the pH will be less than 7 because of the hydrolysis of the conjugate acid.

pH Calculation

The pH is a measure of the hydrogen ion concentration in a solution and is a key indicator of the solution's acidity or basicity. In the context of titrations, it’s particularly important when dealing with weak acids or bases because at the equivalence point, the pH will not be neutral. Rather, it depends on the strength of the weak acid or base involved.

For the titration of triethylamine with \( HCl \) in the exercise, the \( pH \) is calculated using the concentration of \( H^+ \) ions at the equivalence point, which can be found using the \( K_a \) of the conjugate acid of triethylamine as shown in the steps. The \( pH \) is then determined using the negative logarithm of \( [H^+] \) concentration. Recognizing the relationship between \( K_w \) (ion product of water), \( K_b \) of the base, and \( K_a \) of the conjugate acid is crucial for the \( pH \) calculation at the equivalence point. To make the concept clearer for the student, it is helpful to emphasize that \( pK_w = pK_a + pK_b \) and that for a weak base, the \( pH \) at the equivalence point is always greater than 7 due to the presence of the conjugate acid.