Problem 93 - Chemistry The Molecular Science

Question

When \(40.00 \mathrm{~mL}\) of a weak monoprotic acid solution is titrated with \(0.100-\mathrm{M} \mathrm{NaOH}\), the equivalence point is reached when \(35.00 \mathrm{~mL}\) base has been added. After \(20.00 \mathrm{~mL} \mathrm{NaOH}\) solution has been added, the titration mixture has a pH of 5.75. Calculate the ionization constant of the acid.

Step-by-Step Solution

Answer

The ionization constant \(K_a\) of the acid is approximately \(2.37 \times 10^{-6}\).

Step 1: Determine initial moles of NaOH at equivalence point

First, calculate the moles of NaOH added at the equivalence point. Use the volume and concentration of NaOH: \[\text{Moles of } \mathrm{NaOH} = 0.100\,\text{M} \times 0.035\,\text{L} = 0.0035\,\text{moles}\]

Step 2: Equivalence point and acid moles

At the equivalence point, the moles of NaOH equal the moles of the monoprotic acid. Therefore, the moles of the acid is also 0.0035 moles.

Step 3: Calculate moles of NaOH added after 20.00 mL

Next, calculate the moles of NaOH added after the 20.00 mL increment:\[\text{Moles of } \mathrm{NaOH} = 0.100\,\text{M} \times 0.020\,\text{L} = 0.002\,\text{moles}\]

Step 4: Calculate moles of acid remaining and conjugate base formed

Determine the moles of acid remaining and conjugate base formed after 20.00 mL of NaOH has been added:- Initial moles of acid = 0.0035 moles- Moles of acid reacted with 20.00 mL NaOH = 0.002 molesRemaining acid moles: \[0.0035 - 0.002 = 0.0015\,\text{moles}\]Moles of conjugate base formed = 0.002 moles.

Step 5: Calculate concentrations in the titration mixture

The total volume of the solution after adding 20.00 mL NaOH is:\[40.00\,\text{mL} + 20.00\,\text{mL} = 60.00\,\text{mL} = 0.060\,\text{L}\]Concentration of remaining acid:\[\frac{0.0015}{0.060} = 0.025\,\text{M}\]Concentration of conjugate base:\[\frac{0.002}{0.060} = 0.0333\,\text{M}\]

Step 6: Use the Henderson-Hasselbalch equation

Apply the Henderson-Hasselbalch equation to find the ionization constant \(K_a\):\[pH = pK_a + \log\left(\frac{[\text{conjugate base}]}{[\text{acid}]}\right)\]Given \(pH = 5.75\):\[5.75 = pK_a + \log\left(\frac{0.0333}{0.025}\right)\]Calculate the ratio in log:\[\log\left(\frac{0.0333}{0.025}\right) \approx 0.124\]Thus,\[5.75 = pK_a + 0.124\]Solving for \(pK_a\):\[pK_a = 5.75 - 0.124 = 5.626\]

Step 7: Convert pKa to Ka

Use the relationship between \(pK_a\) and \(K_a\) to convert:\[K_a = 10^{-pK_a} = 10^{-5.626} \approx 2.37 \times 10^{-6}\]

Key Concepts

Ionization Constant

The ionization constant, denoted as \(K_a\), is crucial in understanding acid-base reactions. It represents the strength of a weak acid in a solution.
Measured by the dissociation of H⁺ ions in water, \(K_a\) values smaller than 1 indicate that the acid does not fully ionize in solution.
Thus, weak acids with lower \(K_a\) values don't release many protons compared to strong acids.

• For our exercise, we calculated \(K_a\) using values from the titration process.
• We first determined the \(pK_a\) through the Henderson-Hasselbalch equation, then converted it to \(K_a\).
• In this specific titration, the calculated \(K_a\) is about \(2.37 \times 10^{-6}\).

Remember, the lower the \(K_a\), the weaker the acid, with minimal dissociation in water.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation simplifies the calculation of the pH of buffer solutions. It correlates pH, \(pK_a\), and the concentration ratio of acid to its conjugate base.
This powerful equation is \[pH = pK_a + \log\left(\frac{[\text{conjugate base}]}{[\text{acid}]}\right)\]In our exercise, we used this equation to find the \(pK_a\) of the acid after adding a known volume of \(\mathrm{NaOH}\).

• Knowing the pH after a certain amount of titrant helps find the \(pK_a\) of the acid.
• The ratio of conjugate base to acid concentration is crucial in this calculation.

By rearranging this equation, we can deduce the acid's ionization level based on the pH and known concentrations.

Monoprotic Acid

Monoprotic acids are acids that can donate one proton (H⁺) per molecule in a chemical reaction. These acids participate prominently in titration procedures similar to what's described in our exercise.
Because only one proton is donated, calculations during titration become slightly more straightforward.

• In this specific case, the monoprotic nature means that the moles of acid equal the moles of \(\mathrm{NaOH}\) at the equivalence point.
• This characteristic aids in pinpointing when the titration reaction reaches equilibrium.

Monoprotic acids frequently include common acids like acetic acid, hydrochloric acid, and the weak acids used in our example. Their simplicity makes them great learning tools for understanding basic chemistry principles.

Equivalence Point

In titration, the equivalence point is reached when the moles of titrant added equal the moles of substance present in the solution.
It represents the complete neutralization of the acid by the base or vice versa.

• For our task, the equivalence point signified the volume at which moles of \(\mathrm{NaOH}\) equaled moles of the weak monoprotic acid.
• This point was reached when 35.00 mL of \(\mathrm{NaOH}\) was added.

Identifying the equivalence point in titrations helps in understanding the stoichiometry of chemical reactions involved.
In practical applications, finding the equivalence point allows you to accurately determine the concentration of unknown solutions.