Problem 97 - Chemistry

Question

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

Step-by-Step Solution

Answer

The pH of the 0.20 M \(C_2H_5NH_2\) solution is 12.23.

Step 1: Write the base dissociation equation

Ethylamine (C₂H₅NH₂) reacts with water to form the ethylammonium ion (C₂H₅NH₃⁺) and a hydroxide ion (OH⁻). The chemical equation for this reaction is: C₂H₅NH₂ + H₂O ⇌ C₂H₅NH₃⁺ + OH⁻

Step 2: Set up an ICE table

In order to solve for the equilibrium concentration of OH⁻, we'll need to set up an Initial, Change, and Equilibrium (ICE) table for the reaction: ``` [I] 0.20 M - 0 0 [C] -x - +x +x [E] 0.20-x - x x ``` where [I] is the initial concentration, [C] is the amount that changes, and [E] is the equilibrium concentration.

Step 3: Write the Kb expression

In order to find the equilibrium constant for this reaction, we'll need to write an expression for Kb: Kb = \(\frac{[C₂H₅NH₃⁺][OH⁻]}{[C₂H₅NH₂]}\)

Step 4: Substitute the equilibrium concentrations in the Kb expression

Now, we'll substitute the equilibrium concentrations from our ICE table into the Kb expression: \(5.6 × 10^{-4} = \frac{x^2}{0.20-x}\)

Step 5: Solve for the hydroxide ion concentration (x)

Here, we'll assume that x is much smaller than 0.20, meaning that the change in the C₂H₅NH₂ concentration is negligible. This will simplify our reasoning slightly: \(5.6 × 10^{-4} = \frac{x^2}{0.20}\) Now, solve for x: \(x = \sqrt{(5.6 × 10^{-4})(0.20)}\) \(x = 0.0167\) Therefore, [OH⁻] = 0.0167 M.

Step 6: Calculate the pH

To find the pH of the solution, we first need to calculate the pOH using the formula: pOH = -\(\log (0.0167)\) Now, calculate pOH: pOH = 1.77 Finally, we'll use the relationship between pH and pOH (pH + pOH = 14) to find the pH of the solution: pH = 14 - pOH pH = 14 - 1.77 pH = 12.23 The pH of the 0.20 M C₂H₅NH₂ solution is 12.23.

Key Concepts

Base Dissociation Constant

The base dissociation constant, denoted as \( K_b \), is essential in understanding how bases dissociate in water. It describes the extent to which a base can produce hydroxide ions \((\text{OH}^-)\) in solution. A high \( K_b \) value indicates a strong base, meaning it dissociates significantly in solution.

In the exercise, ethylamine \((\text{C}_2\text{H}_5\text{NH}_2)\) reacts with water to form the ethylammonium ion \((\text{C}_2\text{H}_5\text{NH}_3^+)\) and \(\text{OH}^-\). The given \( K_b \) for ethylamine is \(5.6 \times 10^{-4}\), which signifies it is a relatively weak base that doesn't fully dissociate.

Understanding \( K_b \) helps you determine equilibrium concentrations and calculate pH for solutions involving bases.

ICE Table

The ICE table is a useful tool in chemistry to track concentration changes in reactions. ICE stands for Initial, Change, and Equilibrium, allowing you to visualize changes in molarity as the reaction proceeds.

To set up an ICE table for the dissociation of ethylamine:

• Initial: Starting concentration of \( \text{C}_2\text{H}_5\text{NH}_2 \) is 0.20 M, while products begin at 0 M.
• Change: As dissociation occurs, a change of \( -x \) happens for ethylamine and \( +x \) for each product.
• Equilibrium: The final concentration becomes \( 0.20-x \) for ethylamine and \( x \) for the products.

Using an ICE table simplifies visualizing the relationship between initial and equilibrium concentrations, enabling easier calculations of dissociation reactions.

Equilibrium Concentration

Equilibrium concentrations represent the amounts of each species in a reaction mixture at equilibrium. Here, the focus is on finding \( [\text{OH}^-] \) at equilibrium.

Using the dissociation constant expression:
\[ K_b = \frac{[\text{C}_2\text{H}_5\text{NH}_3^+][\text{OH}^-]}{[\text{C}_2\text{H}_5\text{NH}_2]} \]
you substitute values from the ICE table:
\[ 5.6 \times 10^{-4} = \frac{x^2}{0.20-x} \]
To simplify calculations, assume \( x \) is very small compared to 0.20, resulting in:
\( 5.6 \times 10^{-4} = \frac{x^2}{0.20} \)
Solving gives \( x = 0.0167 \) M, the equilibrium concentration of \( [\text{OH}^-] \).

This step is crucial for further pH calculations.

pH and pOH Relationship

The pH and pOH relationship is essential in understanding the acidity and basicity of solutions. They are connected by the equation:

\[ \text{pH} + \text{pOH} = 14 \]

In the exercise, once the \([\text{OH}^-]\) concentration is determined, pOH is calculated as:
\[ \text{pOH} = -\log(0.0167) = 1.77 \]
Using the relationship between pH and pOH, you can then find pH:
\[ \text{pH} = 14 - \text{pOH} = 14 - 1.77 = 12.23 \]

This demonstrates the solution's basic nature, highlighting the importance of understanding how pH and pOH interrelate to characterize chemical solutions.