Question

Aniline hydrochloride, \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}\right) \mathrm{Cl}\), is a weak acid. (Its conjugate base is the weak base aniline, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2} .\) ) The acid can be titrated with a strong base such as NaOH. $$\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell)$$ Assume \(50.0 \mathrm{mL}\) of \(0.100 \mathrm{M}\) aniline hydrochloride is titrated with 0.185 M NaOH. (Ka for aniline hydrochloride is \(2.4 \times 10^{-5} .\) ) (a) What is the pH of the \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}\right)\) Cl solution before the titration begins? (b) What is the \(\mathrm{pH}\) at the equivalence point? (c) What is the pH at the halfway point of the titration? (d) Which indicator in Figure 17.11 could be used to detect the equivalence point? (e) Calculate the pH of the solution after adding \(10.0,20.0,\) and \(30.0 \mathrm{mL}\) of base. (f) Combine the information in parts (a), (b), \((c),\) and \((e),\) and plot an approximate titration curve.

Short answer

(a) 2.81; (b) 8.52; (c) 4.62; (d) Phenolphthalein; (e) 6.01, 4.62, 11.84.

Step-by-step solution

Determine Initial pH of Aniline Hydrochloride Solution

The initial solution is a weak acid solution. Since the substance in question is a weak acid, use the initial concentration and known dissociation constant to calculate the initial pH. The initial concentration is 0.100 M and the acid dissociation constant \( K_a = 2.4 \times 10^{-5} \). For a weak acid \( \mathrm{HA} \), \( [H^+] = \sqrt{K_a \cdot [\text{acid}]} \). Substituting the values, we find:\[[H^+] = \sqrt{2.4 \times 10^{-5} \times 0.100}= \sqrt{2.4 \times 10^{-6}} = 1.55 \times 10^{-3}\]Thus, the initial pH is:\[\text{pH} = -\log[H^+] = -\log(1.55 \times 10^{-3}) = 2.81\]

Determine pH at the Equivalence Point

At the equivalence point, the solution contains only the conjugate base aniline \( \mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2 \). The base dissociation constant \( K_b \) can be found using the relation \( K_w = K_a \cdot K_b \), where \( K_w = 1.0 \times 10^{-14} \).\[K_b = \frac{1.0 \times 10^{-14}}{2.4 \times 10^{-5}} = 4.17 \times 10^{-10}\]The concentration of the conjugate base at equivalence is equal to the initial concentration of the weak acid (0.050 mol in 50 mL is diluted to ~0.027 mol/L in 90 mL, because 50 mL acid + 40 mL base is needed to reach equivalence): Apply the equation for a weak base:\[[OH^-] = \sqrt{0.027 \cdot 4.17 \times 10^{-10}} = 3.33 \times 10^{-6}\]This gives:\[pOH = -\log(3.33 \times 10^{-6}) = 5.48 \Rightarrow \text{pH} = 14 - 5.48 = 8.52\]

Determine pH at Halfway Point

At the halfway point, the concentration of the weak acid \( \text{HA} \) equals the concentration of the conjugate base \( \text{A}^- \). Thus, \( pH = pKa \) because of the buffer equation: \( \text{pH} = pKa + \log\frac{[A^-]}{[HA]} \).So, from the provided \( K_a \):\[pK_a = -\log(2.4 \times 10^{-5}) = 4.62 \]Thus, the pH at the halfway point is 4.62.

Indicator Selection

Select an indicator that changes color around the pH of the equivalence point, which is 8.52. Referring to Figure 17.11 in your textbook (which you should check), the indicator "phenolphthalein" changes color around pH 8.2 to 10, making it a suitable candidate.

Calculate pH After Adding 10.0 mL of Base

When 10 mL of 0.185 M NaOH is added to the 50 mL of 0.100 M aniline hydrochloride solution, 0.00185 moles of base neutralizes 0.001 mole of the acid, turning 0.00085 moles of the remaining acid into its conjugate base form. Total volume is now 60 mL (so concentrations change):- Remaining \( \text{C}_6\text{H}_5\text{NH}_3^+\): \(\frac{0.002 - 0.00185}{0.060} = 0.0025\,\text{M} \) - Formed \( \text{C}_6\text{H}_5\text{NH}_2 \): \(\frac{0.00185}{0.060} = 0.0308\,\text{M}\)Using the buffer equation:\[pH = 4.62 - \log\left(\frac{0.0025}{0.0308}\right) = 4.62 - (-1.39) = 6.01\]

Calculate pH After Adding 20.0 mL of Base

When 20 mL of base is added, 0.0037 moles of NaOH reacts:- Initial moles of aniline hydrochloride = 0.005- Remaining \( \text{C}_6 \text{H}_5 \text{NH}_3^+ \): \(0.0025\,\text{mole}\)- Converted \( \text{C}_6 \text{H}_5 \text{NH}_2 \): \(0.0025\,\text{mole}\)This is exactly the halfway point, and thus:\[pH = \text{pKa} = 4.62\]

Calculate pH After Adding 30.0 mL of Base

Here, 0.00555 moles of NaOH reacts:- Initial moles = 0.005 (neutralization is complete).- Equivalent formation: \(0.0025\,\text{mole}\) \(\text{C}_6 \text{H}_5 \text{NH}_2 \)-An excess \( \text{OH}^- \): \(0.00055\text{mole} \) remaining in 80 mL. Calculating excess pH: \[[OH^-] = \frac{0.00055}{0.080} = 0.006875 \,M\Rightarrow pOH = -\log(0.006875) = 2.16 \Rightarrow pH = 14 - 2.16 = 11.84.\]

Titration Curve Construction

Plot the pH values calculated at different stages—before adding NaOH, at halfway, equivalence, and after specific NaOH volumes. The key points are: (0, 2.81), (50, at equivalence, 8.52), specific additions: (10, 6.01), (20, 4.62), and (30, 11.84). - The curve will rise sharply at the equivalence point from about 7 to 11, reflecting the pH change as the strong base neutralizes the weak acid.

Key concepts

Aniline Hydrochloride

Aniline hydrochloride is a chemical compound that acts as a weak acid in solution. It is structured from aniline, a weak base (\( \mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2 \)), by the addition of a hydrogen ion, converting it to \( \mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_3^+ \). This extra proton makes it capable of donating a proton in a reaction, characterizing its acidic behavior.
When in aqueous solution, aniline hydrochloride partially dissociates, unlike strong acids that dissociate completely. This partial dissociation is because aniline hydrochloride is a weak acid, meaning its molecules do not fully break apart into ions.
Understanding such weak acids is crucial in many chemical reactions, particularly in titrations where the focus is on how these compounds behave when they meet a base.

Weak Acid

Weak acids are substances that do not fully ionize in solution, which differentiates them from strong acids. Because of their partial ionization, the calculation of pH for weak acids involves an equilibrium constant, known as the acid dissociation constant, \( K_a \).
With weak acids, the formula \( [H^+] = \sqrt{K_a \cdot [\text{acid}]} \) is used to determine the concentration of hydrogen ions, which helps to find the pH of the solution. This calculation is evident with aniline hydrochloride, which has a \( K_a \) value that is small, showing weak acidity.
Working with weak acids, such as aniline hydrochloride in a titration, often involves understanding both the initial and the point of equivalence, where the weak acid is entirely transformed into its conjugate base. This transition plays a key role during the titration process.

Equivalence Point

The equivalence point in a titration is the juncture at which the amount of titrant added is precisely enough to neutralize the analyte solution. During the titration of a weak acid with a strong base, like aniline hydrochloride with NaOH, the equivalence point marks where all the weak acid, \( \mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_3^+ \), has been converted to its conjugate base, \( \mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2 \).
At this stage, the solution's pH is influenced by the conjugate base's basic nature. For aniline, the \( K_b \) is calculated using the relation \( K_w = K_a \times K_b \), showing a slight basicity and leading to a resulting pH above 7, usually around 8.52 for aniline.
Recognizing the equivalence point is crucial, as it dictates the selection of a suitable pH indicator for the titration, ensuring accurate detection of this key transition point.

Buffer Solution

A buffer solution is important in maintaining a stable pH even when small amounts of acid or base are added. During a titration involving a weak acid, such as aniline hydrochloride, with a strong base, a buffer solution develops before reaching the equivalence point.
This buffering occurs around the halfway point of the titration, where the amount of acid is equal to the amount of its conjugate base, leading to \( pH = pK_a \). This property helps resist drastic pH changes when additional strong base is added.
Buffer solutions are crucial in many chemical processes, as they help stabilize environments that are sensitive to pH fluctuations. In a practical setting, they allow precise control over the chemical reaction, making them particularly useful in quantitative analysis and various scientific applications.