Question

The \(K_{\mathrm{b}}\) for ethylamine \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NH}_{2}\right)\) is \(4.3 \times 10^{-4}\). What is the \(\mathrm{pH}\) of a \(0.250 \mathrm{M}\) solution of ethylamine? Compare with the \(\mathrm{pH}\) of a \(0.250 \mathrm{M}\) solution of ammonia \(\left(K_{b}=1.86 \times 10^{-5}\right)\).

Short answer

Answer: The pH of the 0.250 M ethylamine solution is 12.01, while the pH of the 0.250 M ammonia solution is 11.33. Ethylamine is the stronger base as its solution has a higher pH.

Step-by-step solution

Write the base dissociation reactions

For ethylamine and ammonia, we can write the base dissociation reactions as follows: For ethylamine: \(CH_3CH_2NH_2 + H_2O \leftrightarrows CH_3CH_2NH_3^+ + OH^-\) For ammonia: \(NH_3 + H_2O \leftrightarrows NH_4^+ + OH^-\)

Set up the \(K_b\) expression and create the reaction table

Consider the initial concentration of ethylamine to be equal to 0.250 M. We can create a table to represent the concentrations as the reaction proceeds: $$\begin{array}{c|c|c|c} & CH_3CH_2NH_2 & + & OH^- & \leftrightarrows & CH_3CH_2NH_3^+ \\ \hline \text{Initial} & 0.250 & & 0 & & 0 \\ \text{Change} & -x & & +x & & +x \\ \text{Equilibrium} & 0.250-x & & x & & x \end{array}$$ The \(K_b\) expression for ethylamine is: \(K_b = \frac{[CH_3CH_2NH_3^+][OH^-]}{[CH_3CH_2NH_2]}\)

Solve for x

Use the given \(K_b\) value to solve for x: \(4.3 \times 10^{-4} = \frac{x^2}{0.250 - x}\) Since \(K_b\) is small, we assume that \(x\) is negligible compared to 0.250. Therefore: \(4.3 \times 10^{-4} = \frac{x^2}{0.250}\) Solving for \(x\) gives the equilibrium concentration of \(OH^-\) ions: \(x = [OH^-] = 0.0103M\)

Calculate the pH

To find the pH of the ethylamine solution, first calculate the pOH using the \(OH^-\) concentration: \(pOH = -\log [OH^-] = -\log(0.0103) = 1.99\) Now, use the relationship between pH and pOH: \(pH = 14 - pOH = 14 - 1.99 = 12.01\) So the pH of the ethylamine solution is 12.01. Now, we repeat Steps 2, 3, and 4 for ammonia. Set up the reaction table and use the given \(K_b\) value to solve for x: $$\begin{array}{c|c|c|c} & NH_3 & + & OH^- & \leftrightarrows & NH_4^+ \\ \hline \text{Initial} & 0.250 & & 0 & & 0 \\ \text{Change} & -x & & +x & & +x \\ \text{Equilibrium} & 0.250-x & & x & & x \end{array}$$ The \(K_b\) expression for ammonia is: \(K_b = \frac{[NH_4^+][OH^-]}{[NH_3]}\) \(1.86 \times 10^{-5} = \frac{x^2}{0.250}\) Solving for x gives the equilibrium concentration of \(OH^-\) ions: \(x = [OH^-] = 0.00216M\) Now, calculate the pOH and then the pH of the ammonia solution: \(pOH = -\log [OH^-] = -\log(0.00216) = 2.67\) \(pH = 14 - pOH = 14 - 2.67 = 11.33\) So the pH of the ammonia solution is 11.33.

Compare the pH values.

The pH of the 0.250 M ethylamine solution is 12.01, while the pH of the 0.250 M ammonia solution is 11.33. Since the \(K_b\) value of ethylamine is larger than that of ammonia, its solution has a higher pH. This indicates that ethylamine is a stronger base than ammonia.

Key concepts

Ethylamine

Ethylamine is an organic compound with the chemical formula CH₃CH₂NH₂. It is a simple aliphatic amine that contains an ethyl group (CH₃CH₂−) attached to an amino group (NH₂). Due to its structure, ethylamine is quite polar.
Acting as a Bronsted-Lowry base, ethylamine can accept a proton from water, forming ethylammonium ions (CH₃CH₂NH₃⁺) and hydroxide ions (OH⁻). This reaction in water increases the pH, showcasing the basic nature of amines due to the lone pair of electrons on the nitrogen atom.

• Usage: Ethylamine is utilized in the synthesis of various medications and chemicals.
• Solubility: It is soluble in water, manifesting as a clear, colorless liquid with a strong ammoniacal odour.
• Basicity: Ethylamine is more basic than ammonia, as evident from its higher base dissociation constant ( K_b g.t.). This means ethylamine dissociates more in solution, making it a stronger base.

Ammonia

Ammonia, with the formula NH₃, is a colorless gas with a pungent odor. It is both highly soluble in water and used in a number of household and industrial applications due to its cleaning properties. When dissolved in water, ammonia acts as a base through a reaction with water, forming ammonium ions (NH₄⁺) and hydroxide ions (OH⁻).

• Chemical Structure: Ammonia consists of a nitrogen atom bonded to three hydrogen atoms, with a lone pair of electrons on nitrogen.
• Basic Characteristics: As a weak base, ammonia partially accepts protons in water, increasing the pH of the solution slightly.
• Industrial Importance: It is widely used both as a fertilizer in agriculture and in the manufacture of explosives, tetraalkyl ammonium salts, and other compounds.

Despite its many applications, safety measures are necessary since ammonia can be corrosive and toxic at high concentrations. Understanding its alkaline nature and behavior in solutions is essential for handling.

Base Dissociation Constant

The Base Dissociation Constant, denoted as K_b, is a crucial concept in understanding the strength of a base. It provides a measure of the tendency of a base to dissociate in water. The larger the value of K_b, the stronger the base, and subsequently, the more OH⁻ ions it produces in an aqueous solution.
For instance, ethylamine has a K_b of 4.3 × 10³, indicating it is a relatively strong base compared to ammonia, which has a K_b of 1.86 × 10⁵.

• Calculation: K_b is derived from the equilibrium concentrations of the base (BOH) and its products (B⁺ and OH⁻) in solution:\[K_b = \frac{[B^+][OH^-]}{[BOH]}\]
• pH Influence: A base with a high K_b increases the pH more significantly, thus indicating higher basic strength.
• Comparison: Comparing K_b values allows us to infer the relative strength of different bases, as seen with ethylamine and ammonia.

Understanding K_b helps in predicting the behavior of base solutions and designing calculations related to pH and equilibrium chemistry. Effective assessment ensures precise applications in both laboratory and industrial settings.