Problem 73 - Chemistry The Central Science

Question

Calculate the molar concentration of \(\mathrm{OH}^{-}\) ions in a \(0.075 \mathrm{M}\) solution of ethylamine \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2} ; K_{b}=6.4 \times 10^{-4}\right) .\) Calcu- late the \(\mathrm{pH}\) of this solution.

Step-by-Step Solution

Answer

The molar concentration of OH- ions in the 0.075 M solution of ethylamine is approximately 0.0068 M, and the pH of the solution is approximately 11.83.

Step 1: Write the balanced chemical equation for the dissociation of the weak base

The ethylamine, C2H5NH2, acts as a weak base and donates a proton to water (H2O) molecules, resulting in the formation of ethylammonium ions (C2H5NH3+) and hydroxide ions (OH-). The balanced chemical equation for this reaction is: C2H5NH2 (aq) + H2O (l) ⇌ C2H5NH3+ (aq) + OH- (aq)

Step 2: Write the expression for the base dissociation constant (Kb)

The base dissociation constant, Kb, relates the equilibrium concentrations of products and reactants for the above reaction. Kb is given as: Kb = \(\frac{[C_{2}H_{5}NH_{3}^{+}][OH^{-}]}{[C_{2}H_{5}NH_{2}]}\)

Step 3: Set up a table for the calculation of equilibrium concentrations

We will call the change in concentration of C2H5NH2, C2H5NH3+, and OH- by x. At equilibrium, we can express their concentrations as: [C2H5NH2] = 0.075 - x [C2H5NH3+] = x [OH-] = x

Step 4: Substitute the expressions and Kb into the equation

Substitute the concentration terms and given Kb value into the equation: \(6.4 \times 10^{-4}\) = \(\frac{x^2}{0.075-x}\)

Step 5: Simplify and solve for x

Since the dissociation of the weak base is minimal, we can assume that x is very small compared to 0.075. So, the equation simplifies to: \(6.4 \times 10^{-4}\) ≈ \(\frac{x^2}{0.075}\) Solving for x, we get: x ≈ \(\sqrt{(6.4 \times 10^{-4})(0.075)}\) x ≈ 0.0068 M

Step 6: Calculate the pOH and pH of the solution

Now that we have the concentration of OH- ions, we can calculate the pOH and then the pH of the solution. The pOH is given by: pOH = -log[OH-] pOH = -log(0.0068) pOH ≈ 2.17 Since pH + pOH = 14, we can find the pH: pH = 14 - pOH pH = 14 - 2.17 pH ≈ 11.83 The molar concentration of OH- ions in the 0.075 M solution of ethylamine is approximately 0.0068 M, and the pH of the solution is approximately 11.83.

Key Concepts

Weak Base

A weak base, like ethylamine (\(\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{NH}_{2}\)), essentially participates in an incomplete proton transfer reaction with water. This means that only a small fraction of the base molecules accept \(\mathrm{H}^+\) ions from water.
The resulting ions are the conjugate base, ethylammonium (\(\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{NH}_{3}^+\)), and hydroxide ions (\(\mathrm{OH}^-\)).
When dissolved in water, weak bases do not break apart as completely as strong bases do, leaving a notable amount of undissociated base in the solution:

• Weak base example: \(\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{NH}_{2} + \mathrm{H}_{2}\mathrm{O} \leftrightarrow \mathrm{C}_{2}\mathrm{H}_{5}\mathrm{NH}_{3}^{+} + \mathrm{OH}^{-}\)
• This equilibrium is crucial as it determines the balance between the base and the ions it produces.

Highlighting this balance forms the basis for understanding base behavior in solutions.

Base Dissociation Constant

The base dissociation constant (\(K_b\)) provides valuable insight into the extent to which a base dissociates in water.
For ethylamine, the value \(K_b = 6.4 \times 10^{-4}\) indicates a relatively weak dissociation.

The equation relating \(K_b\) to concentrations at equilibrium is:

• \[K_b = \frac{\left[\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{NH}_{3}^{+}\right]\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{NH}_{2}\right]}\]

The smaller the \(K_b\) value, the weaker the base and the less it dissociates:

• A higher concentration of undissociated base indicates weak ionization.
• Knowing \(K_b\) helps predict how much base remains unreacted.

Understanding \(K_b\) is essential in pH calculation and determining ion concentrations in solutions.

Equilibrium Concentrations

Equilibrium concentrations represent the final concentrations of reactants and products in a chemical equilibrium.
In our case, it involves the concentrations of ethylamine, ethylammonium ions, and hydroxide ions.

At equilibrium:

• Start concentration of ethylamine: 0.075 M
• Concentration of \(\mathrm{OH}^-\) and \(\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{NH}_{3}^+\) ions formed: \(x\)
• Remaining concentration of ethylamine: \(0.075 - x\)

Due to the weak base's minimal dissociation, \(x\) is often small:

• In approximations, \(x\) is sometimes neglected compared to larger values.
• This simplification helps solve for \(x\) mathematically quicker and predict values efficiently.

Grasping equilibrium concentrations aids in understanding the dynamics and chemical balance of a reaction.

pH Calculation

Calculating \(pH\) involves understanding the relationship between \(\mathrm{pOH}\) and \(\mathrm{pH}\), where: \[\mathrm{pH} + \mathrm{pOH} = 14\]
To find the \(\mathrm{pH}\) of a solution with a known concentration of hydroxide ions (\(\mathrm{OH}^{-}\)), we first calculate the \(\mathrm{pOH}\):

• \[\mathrm{pOH} = -\log[\mathrm{OH}^{-}]\]
• For [\mathrm{OH}^{-}] = 0.0068 M,\[\mathrm{pOH} \approx 2.17\]
• Using the formula, \[\mathrm{pH} = 14 - \mathrm{pOH}\], of approximately \[11.83\] is calculated.

This indicates a basic solution, typical for weak bases, as they increase the concentration of hydroxide ions. By converting \(\mathrm{OH}^{-}\) to pH, one measures the acidity or basicity of the solution directly.