Problem 86 - Chemistry: The Central Science in SI Units, Global Edition

Question

The acid-base indicator bromcresol green is a weak acid. The yellow acid and blue base forms of the indicator are present in equal concentrations in a solution when the pH is 4.68 . What is the \(\mathrm{p} K_{a}\) for bromcresol green?

Step-by-Step Solution

Answer

The \(\mathrm{p} K_{a}\) for bromcresol green is 4.68, as determined by using the Henderson-Hasselbalch equation with the given pH and equal concentrations of the acid and base forms of the indicator.

Step 1: Write down the Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is as follows: \(\mathrm{pH} = \mathrm{p} K_{a} + \log_{10} \frac{[\mathrm{A}^{-}]}{[\mathrm{HA}]}\) where \(\mathrm{pH}\) is the pH of the solution, \(\mathrm{p} K_{a}\) is the pKa of the weak acid, \([\mathrm{A}^{-}]\) is the concentration of the conjugate base, and \([\mathrm{HA}]\) is the concentration of the weak acid. In this exercise, the acid and base forms of bromcresol green are present in equal concentrations, so \([\mathrm{A}^{-}] = [\mathrm{HA}]\).

Step 2: Set up the equation with the given values

Now, let's plug in the given values into the Henderson-Hasselbalch equation: \(\mathrm{pH} = \mathrm{p} K_{a} + \log_{10} \frac{[\mathrm{A}^{-}]}{[\mathrm{HA}]}\) \(4.68 = \mathrm{p} K_{a} + \log_{10} \frac{[\mathrm{A}^{-}]}{[\mathrm{HA}]}\) Since \([\mathrm{A}^{-}] = [\mathrm{HA}]\), the fraction in the equation becomes: \(\frac{[\mathrm{A}^{-}]}{[\mathrm{HA}]} = \frac{[\mathrm{HA}]}{[\mathrm{HA}]} = 1\)

Step 3: Calculate the pKa of bromcresol green

Now the equation becomes: \(4.68 = \mathrm{p} K_{a} + \log_{10}(1)\) Since \(\log_{10}(1) = 0\), we have: \(4.68 = \mathrm{p} K_{a}\) So the \(\mathrm{p} K_{a}\) for bromcresol green is 4.68.

Key Concepts

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a useful mathematical tool in acid-base chemistry. It helps us relate the pH of a solution to the pKa of a weak acid and the concentrations of the acid and its conjugate base. The equation is given by:

• \( \text{pH} = \text{p}K_a + \log_{10} \left( \frac{[\mathrm{A}^-]}{[\mathrm{HA}]} \right) \)

This formula is especially handy in buffer solutions and finding the pH when acid and base forms are not equal.
When the concentrations of the acid \([\mathrm{HA}]\) and its conjugate base \([\mathrm{A}^-]\) are equal, the logarithmic term becomes zero because \( \log_{10}(1) = 0 \).
Thus, under these conditions, the pH is equal to the pKa of the acid. This brings a considerable simplification because it allows for a straightforward conversion from pH to pKa and vice versa, especially in cases involving indicators like bromcresol green.
Understanding this concept is crucial for predicting how buffers will respond to changes in concentration and what pH range they can stabilize.

pKa Calculation

To find the pKa value of a weak acid, you often start with the Henderson-Hasselbalch equation. When the acid and its conjugate base are present in equal concentrations, the equation simplifies greatly. This scenario occurs at the midpoint of a titration or when dealing with certain indicators.

• Given the equation \( \mathrm{pH} = \mathrm{p}K_a + \log_{10} \left( \frac{[\mathrm{A}^-]}{[\mathrm{HA}]} \right) \), if the concentrations are equal, the ratio \( \frac{[\mathrm{A}^-]}{[\mathrm{HA}]} = 1 \).
• This makes \( \log_{10}(1) = 0 \), simplifying the equation to \( \mathrm{pH} = \mathrm{p}K_a \).

In our exercise involving bromcresol green, the pH provided is 4.68, indicating that the is also 4.68 when the indicator's acid and base forms are equal. This kind of problem can often seem intimidating, but it's a simple plug-and-chug once you understand why the equation simplifies the way it does.

Indicators in Chemistry

Indicators are substances that change color at a particular pH range, helping us identify the end point of a titration. They are typically weak acids or bases themselves.
When an indicator like bromcresol green exists in different color forms in a solution, it is used to visually indicate the pH of the solution.

• The color change occurs due to the equilibrium between the acid form and the base form.
• In our example, at a pH of 4.68, the indicator is at its equivalence point, showing a transition between its acid (yellow) and base (blue) forms.

Such indicators are tailored to specific pH ranges. Therefore, choosing the right indicator depends on the range in which you expect your solution's pH to fall.
Understanding these concepts helps significantly in various chemical applications, from laboratory experiments to industry and beyond. By knowing how indicators work, you can ensure the accuracy and reliability of your analytical results.