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Acid-base indicators are compounds that change color according to the pH of the solution they are in. These indicators are typically weak acids or bases. When dissolved in water, they exhibit different colors based on their protonated (acidic) or deprotonated (basic) forms.
An easy way to remember the behavior of acid-base indicators is that:

• In acidic solutions, the protonated form dominates, often showing a specific color (e.g., red for some indicators).
• In basic solutions, the deprotonated form prevails, displaying another color (e.g., blue).

To exemplify this, imagine a solution which appears red when acidic and blue when basic. By observing the color change, you can estimate the pH of the solution. As indicated in the exercise, the shift from 75% red (acidic) to 75% blue (basic) involves the use of the Henderson-Hasselbalch equation to determine the pH change necessary for this transition.

The pKa value of an acid-base indicator is crucial to understanding how it functions. pKa simplifies the logarithmic acid dissociation constant \(K_a\). This offers an easy way to see how easily the compound donates protons. The calculation for pKa from \(K_a\) is made using the formula:\[\text{pKa} = -\log(\mathrm{K}_a)\]This number signifies the pH at which the indicator has equal concentrations of both its acidic and basic forms.
For instance, \(K_a = 3 \times 10^{-5}\) leads to:\[\text{pKa} = -\log(3 \times 10^{-5}) = 4.523\]This pKa implies that at pH 4.523, the indicator's color would be halfway between its acidic and basic color forms. Knowing the pKa allows us to use the Henderson-Hasselbalch equation to predict color changes at varying pH levels accurately.

pH change is central when considering acid-base indicators because it directly influences which form of the indicator is dominant. The Henderson-Hasselbalch equation assists in calculating these shifts. The equation is: \[\text{pH} = \text{pKa} + \log\left(\frac{[\text{Base Form}]}{[\text{Acid Form}]}\right)\]When 75% of the indicator is in its acid form, the solution is primarily colored red. Thus, \(\frac{[\text{Base}]}{[\text{Acid}]} = \frac{1}{3}\).
We calculate the initial pH as follows:\[\text{pH}_{\text{initial}} = 4.523 + \log\left(\frac{1}{3}\right) = 4.046\]For a shift to 75% blue, the \(\frac{[\text{Base}]}{[\text{Acid}]} = 3\). Hence, the pH becomes:\[\text{pH}_{\text{final}} = 4.523 + \log(3) = 5.000\]The pH change \(\Delta \text{pH}\) is the difference between final and initial values. Thus, \[\Delta \text{pH} = 5.000 - 4.046 = 0.954\].This shows a significant enough alteration to complete the color transition as required by the exercise.