Question

Calculate the \(\mathrm{pH}\) of a \(0.20-M \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\) solution \(\left(K_{\mathrm{b}}=\right.\) \(5.6 \times 10^{-4} )\)

Short answer

The pH of a 0.20 M ethylamine solution can be calculated using the given \(K_b= 5.6 \times 10^{-4}\). First, set up an ICE table and the Kb expression, then solve for the concentration of OH- ions, which is approximately 0.0167 M. Calculate the pOH, which is approximately 1.78, and finally, find the pH using the equation pH + pOH = 14. The pH of the solution is approximately 12.22.

Step-by-step solution

Write the balanced equation for the dissociation of ethylamine in water

Initially, we need to write the balanced chemical equation. Ethylamine (C2H5NH2) reacts with water (H2O) to produce ethylammonium ions (C2H5NH3+) and hydroxide ions (OH-). The equation is: \[C2H5NH2 + H2O \longleftrightarrow C2H5NH3^+ + OH^-\]

Set up the ICE table and the Kb expression

Next, we set up an ICE (Initial, Change, Equilibrium) table using the initial concentration of ethylamine given, which is 0.20 M. We can then use the table to write the expression for Kb: - At the start of the reaction: [C2H5NH2] = 0.20 M, [C2H5NH3+] = 0 M, [OH-] = 0 M - At equilibrium: [C2H5NH2] = 0.20 - x, [C2H5NH3+] = x, [OH-] = x The expression for Kb is: \[K_b = \frac{[C2H5NH3^+][OH^-]}{[C2H5NH2]}\]

Solve for x, which is the concentration of OH-.

Now we need to solve for x by plugging in the given Kb and the equilibrium concentrations from the ICE table: \(5.6 \times 10^{-4} = \frac{x^2}{0.20 - x}\) To solve this equation, we can use the approximation \(0.20 - x \approx 0.20\), since x is expected to be very small compared to the initial concentration 0.20 M. This will simplify the equation to: \(5.6 \times 10^{-4} \approx \frac{x^2}{0.20}\) \(x \approx \sqrt{(5.6 \times 10^{-4})(0.20)}\) \(x \approx 0.0167\) The concentration of hydroxide ions is approximately equal to the value of x, which is 0.0167 M.

Calculate the pOH and pH of the solution

Now that we have the concentration of OH-, we can calculate the pOH and pH of the solution. To find the pOH, use the formula: \[pOH = -\log{[OH^{-}]}\] So, \[pOH = -\log{(0.0167)} \approx 1.78\] Finally, to find the pH, we use the equation: \[pH + pOH = 14\] Thus, \[pH = 14 - pOH \approx 12.22\] The pH of a 0.20 M ethylamine solution is approximately 12.22.

Key concepts

Understanding Ethylamine

Ethylamine is an organic compound that belongs to the family of amines. Its chemical formula is \( \mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2 \). This molecule consists of an ethyl group \( \mathrm{C}_2 \mathrm{H}_5 \) attached to an amino group \( \mathrm{NH}_2 \). Amines like ethylamine act as bases because they can accept hydrogen ions from substances like water.

When ethylamine is dissolved in water, it increases the solution's pH by producing hydroxide ions \( \mathrm{OH}^- \) as a result of its basic nature. This process forms part of the chemical equilibrium that we evaluate when calculating the pH of solutions containing amines.

The ICE Table explained

In chemistry, an ICE table is a systematic way of tracking the concentrations of species in a reaction at different stages: Initial, Change, and Equilibrium—hence the acronym ICE. Let's see how it applies to the dissociation of ethylamine in water.

• **Initial stage**: The known concentration, which is 0.20 M for ethylamine, and 0 M for the products.
• **Change stage**: We let the change in concentration be \(x\), representing how much ethylamine reacts or dissociates to form products.
• **Equilibrium stage**: Calculate theoretical concentrations, assuming the equilibrium constant \( K_b \) guides the reaction.

With an ICE table, students can visualize how concentrations grow or shrink and use this information for further calculations.

The Kb Expression's Role

The equilibrium constant for bases, known as \( K_b \), provides the relationship between the concentrations of reactants and products at equilibrium in a basic solution. For the reaction of ethylamine, its \( K_b \) value is \( 5.6 \times 10^{-4} \).

The expression of the base constant is formulated as follows:
\[ K_b = \frac{[\mathrm{C}_2\mathrm{H}_5\mathrm{NH}_3^+][\mathrm{OH}^-]}{[\mathrm{C}_2\mathrm{H}_5\mathrm{NH}_2]} \]
This relationship helps us connect the dots from the ICE table and solve for the unknowns (post-dissociation concentrations), essential for further pH calculations.

Exploring Chemical Equilibrium

Chemical equilibrium is the state in which the concentrations of the reactants and products are constant over time. In the context of ethylamine in water, equilibrium is achieved when the concentration changes from the reaction offset by the formation of products: ethylammonium ions and hydroxide ions.

Understanding equilibrium involves recognizing that reactions can proceed forwards and backwards until a balance is reached. This knowledge is fundamental when calculating pH, especially in weak base reactions like our current scenario with ethylamine.

Hydroxide Ions and pH

Hydroxide ions \(\mathrm{OH}^-\) are crucial in determining a solution's basicity. In our ethylamine example, their concentration determines how basic the solution becomes. As ethylamine dissolves in water, it generates hydroxide ions, which increase the pH value.

To relate hydroxide concentration to pH, we first calculate the pOH using the formula \(\mathrm{pOH} = -\log[\mathrm{OH}^-]\). Then, since \( \mathrm{pH} + \mathrm{pOH} = 14 \), we find the pH, illustrating how increased \(\mathrm{OH}^-\) raises pH levels, confirming the solution's alkalinity.