Problem 20 - General Chemistry

Question

If \(20.0 \mathrm{~mL}\) of a \(0.10 \mathrm{M} \mathrm{NaOH}\) solution is added to a 30.0 -mL sample of a \(0.10 M\) weak acid, HA, what is the \(\mathrm{pH}\) of the resulting solution? \(\left(K_{a}=1.8 \times 10^{-5}\right.\) for \(\left.\mathrm{HA}\right)\) a. 2.87 b. 2.74 c. 4.74 d. 5.05 e. 8.73

Step-by-Step Solution

Answer

The pH is 5.05.

Step 1: Convert Volumes to Liters

First, convert the volumes from milliliters to liters. The volume of the NaOH solution is 20.0 mL, which is equivalent to 0.020 L. Similarly, the volume of the HA solution is 30.0 mL, equivalent to 0.030 L.

Step 2: Calculate Moles of NaOH and HA

Calculate the moles of NaOH and HA in their respective solutions using the formula \( ext{moles} = ext{molarity} \times ext{volume in liters} \).\[ ext{Moles of NaOH} = 0.10 ext{ M} \times 0.020 ext{ L} = 0.002 ext{ moles} \]\[ ext{Moles of HA} = 0.10 ext{ M} \times 0.030 ext{ L} = 0.003 ext{ moles} \]

Step 3: Determine Reaction Completion

In this neutralization reaction, NaOH reacts with HA to form water and the anion A-. Use the equation for the reaction: \( \text{HA} + \text{NaOH} \rightarrow \text{A}^- + \text{H}_2\text{O} \). Since both are 0.10 M solutions, the NaOH will neutralize 0.002 moles of HA, leaving 0.001 moles of HA unreacted (0.003 - 0.002 = 0.001 moles). This results in 0.002 moles of A⁻, the conjugate base.

Step 4: Set Up the Henderson-Hasselbalch Equation

Use the Henderson-Hasselbalch equation to find the pH: \[ \text{pH} = \text{pKa} + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \]Calculate the \( \text{pKa} \) from \( \text{Ka} = 1.8 \times 10^{-5} \) using \( \text{pKa} = -\log(\text{Ka}) = -\log(1.8 \times 10^{-5}) = 4.74 \).

Step 5: Plug in Concentrations into the Equation

Next, calculate the concentration of the weak acid (HA) and its conjugate base (A⁻) in the mixture.\[ \text{[HA]} = \frac{0.001}{0.050} \text{ M} = 0.020 \text{ M} \]\[ \text{[A}^-] = \frac{0.002}{0.050} \text{ M} = 0.040 \text{ M} \]Plug the concentrations into the equation:\[ \text{pH} = 4.74 + \log \left( \frac{0.040}{0.020} \right) = 4.74 + \log(2) = 5.05 \]

Step 6: Final Answer

The pH of the resulting solution is approximately 5.05.

Key Concepts

Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is an incredibly useful tool in the world of chemistry, particularly for acid-base titrations. It allows us to relate the pH of a solution to the concentration of an acid and its conjugate base. The equation is written as follows:
\[\text{pH} = \text{pKa} + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right)\]
Here, \(\text{pKa}\) is the negative logarithm of the acid dissociation constant, \(\text{Ka}\), of the weak acid. It provides a measure of the strength of the acid. The term \([\text{A}^-]\) refers to the concentration of the conjugate base, and \([\text{HA}]\) is the concentration of the remaining weak acid. This equation becomes especially handy when you know the initial concentrations or when a neutralization reaction has occurred, and you need to calculate the pH of the resulting solution. Through this simple equation, understanding the behavior of weak acids in buffered solutions becomes much easier for students and chemists alike.

weak acid

A weak acid is an acid that does not completely dissociate in water. This means that, in a solution of a weak acid, only a small fraction of the acid molecules donate their protons (\(\text{H}^+\)) to the water to form hydronium ions (\(\text{H}_3\text{O}^+\)). Because of this partial ionization, weak acids have a specific characteristic, their \(\text{Ka}\) value, which indicates their ionization potential.
A weak acid like HA in our exercise might only donate a small portion of its protons to a base like NaOH. The reactions between weak acids and bases lead to an equilibrium state in the mixture, where a portion of the weak acid remains as the undissociated form while some is converted into its conjugate base (\(\text{A}^-\)).

• Examples: Acetic acid, citric acid, and carbonic acid are common weak acids.
• Characteristic: Weak acids often have a sour taste and are less electrically conductive than strong acids.
• Importance: They are widespread in biological systems and play key roles in buffering solutions.

pH calculation

Calculating the pH of a solution is an essential skill in chemistry that reveals the acidity or alkalinity of that solution. The pH is expressed as the negative logarithm of the hydrogen ion concentration:
\[\text{pH} = -\log[\text{H}^+ ]\]
In scenarios involving a weak acid and its conjugate base—particularly in the aftermath of a neutralization reaction—the Henderson-Hasselbalch equation can further aid in determining the resulting pH. The procedure typically involves:

• Determining the moles of weak acid and its conjugate base after the reaction.
• Calculating their concentrations by dividing by the total volume of the solution.
• Plugging these values into the Henderson-Hasselbalch equation.

This systematic process ensures that we achieve an accurate pH measurement that reflects how the acid and base interact within the solution. Understanding the principles behind pH calculations helps demystify acid-base reactions and their impacts in both laboratory settings and the natural world.