Problem 158 - A Complete Resource Book in Chemistry for JEE Main

Question

An acid base indicator has \(\mathrm{K}_{\mathrm{a}}=3 \times 10^{-5} .\) The acid form of the indicator is red and the basic form is blue. By how much must the \(\mathrm{pH}\) change in order to change the indicator from \(75 \%\) red to \(75 \%\) blue \((\log 3=0.4770)\) (a) \(0.95\) (b) \(2.3\) (c) \(0.75\) (d) 5

Step-by-Step Solution

Answer

The pH must change by 0.95.

Step 1: Understanding the Problem

We are asked to find how much the pH needs to change to switch the indicator from 75% red to 75% blue. The acid form is red, and the basic form is blue. We need to apply the concept of the Henderson-Hasselbalch equation to solve this.

Step 2: Write the Henderson-Hasselbalch Equation

The equation in the context of an indicator that changes color is: \[ \text{pH} = \text{pKa} + \log\left(\frac{[\text{Base Form}]}{[\text{Acid Form}]}\right)\]Given that \(\mathrm{K}_a = 3 \times 10^{-5}\), we can find \(\text{pKa}\) as follows: \[\text{pKa} = -\log(\mathrm{K}_a)\]

Step 3: Calculate the pKa

Use the provided log value: \[\text{pKa} = -\log(3 \times 10^{-5}) = -\log(3) - \log(10^{-5})\]\[\text{pKa} = -0.4770 + 5 = 4.523\]

Step 4: Calculate Initial pH for 75% Red

When 75% of the indicator is red (acidic form), 25% is in blue (basic form). The ratio \([\text{Base}]/[\text{Acid}] = 0.25/0.75 = 1/3\). Applying this in the Henderson-Hasselbalch equation:\[\text{pH}_{\text{initial}} = 4.523 + \log\left(\frac{1}{3}\right)\]Using \(\log(1/3) = -\log(3) = -0.4770\):\[\text{pH}_{\text{initial}} = 4.523 - 0.477 = 4.046\]

Step 5: Calculate Final pH for 75% Blue

When 75% of the indicator is blue (basic form), 25% is red (acidic form). The ratio \([\text{Base}]/[\text{Acid}] = 0.75/0.25 = 3\). Plug it in:\[\text{pH}_{\text{final}} = 4.523 + \log(3)\]Here \(\log(3) = 0.4770\): \[\text{pH}_{\text{final}} = 4.523 + 0.477 = 5.000\]

Step 6: Determine the pH Change

Find the difference between the final and initial pH values:\[\Delta \text{pH} = \text{pH}_{\text{final}} - \text{pH}_{\text{initial}} = 5.000 - 4.046 = 0.954\]Round to the nearest value from the given options: \(\Delta \text{pH} = 0.95\).

Key Concepts

Acid-Base Indicator

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.

pKa Calculation

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

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.