Question

When aniline, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\left(K_{\mathrm{b}}=7.4 \times 10^{-10}\right)\), reacts with a strong acid, its conjugate acid, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}{ }^{+},\) is formed. Calculate the \(\mathrm{pH}\) of a \(0.100 \mathrm{M}\) solution of \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}{ }^{+}\) and \(\mathrm{com}^{-}\) pare it with the \(\mathrm{pH}\) of acetic acid \(\left(K_{\mathrm{a}}=1.86 \times 10^{-5}\right)\).

Short answer

Answer: The pH of the 0.100 M solution of C6H5NH3+ is approximately 2.94, and it is lower than that of acetic acid.

Step-by-step solution

Calculate the Ka of the conjugate acid

First, we need to find the Ka of the conjugate acid, C6H5NH3+. Since we are given the Kb of aniline (C6H5NH2), we can use the relationship between Ka, Kb, and Kw (the ion-product constant for water): \(K_{a} \times K_{b} = K_{w}\) where Kw = 1.0 x 10^-14 at 25°C. Thus, \(K_{a} = \frac{K_{w}}{K_{b}}\) Plugging in the values for Kb and Kw: \(K_{a} = \frac{1.0 \times 10^{-14}}{7.4 \times 10^{-10}} = 1.35 \times 10^{-5}\)

Use Ka Expression

Now, we will use the Ka expression, which is given by: $K_{a} = \frac{[\mathrm{H}^{+}][\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}]}{[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}{ }^{+}]}$

Write the Expression for the Concentration Changes

Let x moles of H+ dissociate from C6H5NH3+: At equilibrium, \([\mathrm{H}^{+}] = x, [\mathrm{C}_{6} \mathrm{H}_{5}\mathrm{NH}_{2}] = x\) Since the initial concentration of C6H5NH3+ is given as 0.100 M, \( [\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}{ }^{+}] = 0.100 - x \)

Substitute the Equilibrium Concentrations in Ka Expression

Substitute the expressions of equilibrium concentrations in the Ka expression: \(1.35 \times 10^{-5} = \frac{ x^2 }{ (0.100 - x)}\) Assuming x is very small compared to 0.100, we can ignore x in the denominator: \(1.35 \times 10^{-5} \approx \frac{x^2}{0.100}\)

Solve for Concentration of H+ ions

Now, we solve for x: \(x^2 = 1.35 \times 10^{-5} \times 0.100\) \(x^2 = 1.35 \times 10^{-6}\) \(x \approx 1.16 \times 10^{-3} \mathrm{M}\) The concentration of H+ ions is approximately 1.16 x 10^-3 M.

Calculate pH of the Conjugate Acid Solution

Now, using the concentration of H+ ions, we can calculate the pH of the conjugate acid solution: \(pH = -\log{[\mathrm{H}^{+}]}\) \(pH = -\log{(1.16 \times 10^{-3})}\) \(pH \approx 2.94\) The pH of the 0.100 M solution of C6H5NH3+ is approximately 2.94.

Compare with the pH of Acetic Acid

We are also given the Ka of acetic acid, which is 1.86 x 10^-5. To find the pH of acetic acid, we can follow the same process we did for the conjugate acid solution. However, since we are asked to compare the pH values, we can use the Ka values to make a comparison: Since Ka(C6H5NH3+) = 1.35 x 10^-5 > Ka(Acetic Acid) = 1.86 x 10^-5, the pH of C6H5NH3+ will be lower than that of acetic acid, as stronger acids have lower pH values. Thus, the pH of the 0.100 M solution of C6H5NH3+ (approximately 2.94) is lower than that of acetic acid.

Key concepts

Acid-Base Equilibrium

In chemistry, the concept of acid-base equilibrium is pivotal when understanding reactions between acids and bases. This equilibrium involves the transfer of protons (H⁺ ions) from acids to bases. When an acid donates a proton, it becomes its conjugate base, while the recipient base accepting the proton becomes its conjugate acid.

This exchange can be quantified by an equilibrium constant, denoted as Ka for acids, which measures the strength of the acid. A higher Ka value indicates a stronger acid as it demonstrates a greater tendency to donate protons. Conversely, the equilibrium constant for bases is denoted as Kb, and a higher Kb implies a stronger base, capable of accepting more protons. The relationship between Ka and Kb is inversely proportional and is linked via the ion-product constant of water, Kw, which is always 1.0 x 10^-14 at 25°C.

Ka Expression

The Ka expression is a mathematical representation of the acid dissociation constant, a quantitative measure of an acid's strength in solution. It is typically presented as:
\[ K_a = \frac{[\mathrm{H}^{+}][\mathrm{A}^{-}]}{[\mathrm{HA}]} \]
Where:

• \(K_a\) is the acid dissociation constant.
• \([\mathrm{H}^{+}]\) is the concentration of hydrogen ions in the solution.
• \([\mathrm{A}^{-}]\) is the concentration of the conjugate base after dissociation.
• \([\mathrm{HA}]\) is the concentration of the acid before dissociation.

For weak acids, which only partially dissociate, this equilibrium constant is essential for calculating the pH of their solutions. By solving the Ka expression for the concentration of hydrogen ions (\([\mathrm{H}^{+}]\)), one can then apply the negative logarithm to find the pH, exemplifying the ties between Ka, pH, and acid strength.

Conjugate Acids and Bases

When discussing acids and bases, it is vital to understand the role of conjugate pairs. Each acid is linked to a conjugate base, which is formed when the acid donates a proton, and conversely, each base has a conjugate acid, formed when it accepts a proton. For instance, in the reaction involving aniline (\(\mathrm{C}_6 \mathrm{H}_5\mathrm{NH}_2\)), the aniline is a base and its conjugate acid is \(\mathrm{C}_6 \mathrm{H}_5\mathrm{NH}_3^{+}\).

The conjugate acid-base pair concept reflects the reversible nature of acid-base chemistry. Notably, the strength of a conjugate base is inversely related to the acid strength from which it is derived. Strong acids have weak conjugate bases, while weak acids have relatively stronger conjugate bases. This relationship is essential when comparing the acidity of different solutions. By calculating the pH values as shown in our textbook example, we can discern that the solution of \(\mathrm{C}_6 \mathrm{H}_5\mathrm{NH}_3^{+}\), with a calculated pH of approximately 2.94, is more acidic than its weaker acid counterpart, acetic acid (due to a lower Ka value of the latter).