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

0.95

Step 1: Determine Fraction of Color

The indicator changes from 75% red to 75% blue. The red form is the acidic form and the blue form is the basic form. When 75% is in one form, 25% is in the other form. Initially, the ratio of red to blue is \( \frac{3}{1} \), and finally, it becomes blue to red as \( \frac{3}{1} \).

Step 2: Use Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is: \[\text{pH} = \text{pK}_a + \log\left(\frac{\text{[Base]}}{\text{[Acid]}}\right)\]. Here, \(\text{pK}_a = -\log(K_a) = -\log(3 \times 10^{-5})\).

Step 3: Calculate Initial pH

We calculate the initial pH when the indicator is 75% red. Thus, \( \text{[Base]} = \frac{1}{4}\) and \( \text{[Acid]} = \frac{3}{4}\). Use the equation: \[ \text{pH}_{\text{initial}} = \text{pK}_a + \log\left(\frac{1}{3}\right) \] Substitute \(\text{pK}_a = 5 - \log(3) = 5 - 0.4770 = 4.523\). Thus: \[ \text{pH}_{\text{initial}} = 4.523 + \log(\frac{1}{3}) = 4.523 - 0.4770 \approx 4.046 \]

Step 4: Calculate Final pH

We calculate the final pH when the indicator is 75% blue. Thus, \( \text{[Base]} = \frac{3}{4}\) and \( \text{[Acid]} = \frac{1}{4} \). Use the equation: \[ \text{pH}_{\text{final}} = \text{pK}_a + \log\left(\frac{3}{1}\right) \] Thus: \[ \text{pH}_{\text{final}} = 4.523 + \log(3) = 4.523 + 0.477 \approx 5.000 \]

Step 5: Determine Change in pH

Subtract the initial pH from the final pH to find the change in pH: \[ \Delta\text{pH} = \text{pH}_{\text{final}} - \text{pH}_{\text{initial}} = 5.000 - 4.046 \approx 0.954 \]

Step 6: Conclusion

The change in pH required to shift the indicator from 75% red to 75% blue is approximately 0.95.

Key Concepts

Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is a useful tool for calculating the pH of a solution containing a conjugate acid-base pair. It is derived from the acid dissociation constant (Ka) of a weak acid and allows us to estimate the pH based on the relative concentrations of the acid and its conjugate base. This equation is given as: \[\text{pH} = \text{pK}_a + \log\left(\frac{\text{[Base]}}{\text{[Acid]}}\right)\]In this formula, \(\text{pK}_a\) is the negative logarithm of the acid dissociation constant \(K_a\), highlighting the relationship between the strength of an acid and its acidity (pH). When the concentrations of the base and acid are equal, the pH of the solution equals \(\text{pK}_a\), making this equation particularly insightful in buffer solutions.The equation is valuable in practical applications, like when predicting the behavior of pH-sensitive indicators, as seen in this problem where we track the color change of an acid-base indicator from red to blue based on the pH level.

pH calculation

Calculating pH is a crucial part of understanding acid-base chemistry. The pH scale ranges from 0 to 14 and measures the acidity or basicity of a solution. A pH of 7 is neutral, anything less is acidic, and anything more is basic. In the exercise, we use the Henderson-Hasselbalch equation for calculating initial and final pH values during a color transition of an indicator. Initially, when the indicator is 75% red, we find the initial pH. For this, the ratio of the concentrations [Base]:[Acid] is 1:3, thus using the equation, \[ \text{pH}_{\text{initial}} = \text{pK}_a + \log\left(\frac{1}{3}\right) \]Similarly, when 75% of the indicator turns blue, the ratio changes to 3:1, and the final pH is calculated by:\[ \text{pH}_{\text{final}} = \text{pK}_a + \log(3) \]These calculations provide insightful observations on how the pH affects the indicator's color. Understanding the change in pH, \(\Delta\text{pH} = 0.954\), gives us a clear picture of the sensitivity of the indicator to pH changes.

equilibrium constant (Ka)

The equilibrium constant for an acid, denoted as \(K_a\), quantifies the strength of an acid in solution. It represents the acid's ability to donate protons to the base, a vital concept in acid-base chemistry. Mathematically, it is expressed as the ratio of the concentrations of the products (conjugate base and the hydronium ion) to the reactants (acid).\[\text{K}_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}\]A smaller \(K_a\) value indicates a weaker acid, as less of the acid dissociates to form the corresponding ions. In the given problem, the \(K_a = 3 \times 10^{-5}\), which tells us the indicator is a weak acid. The \(\text{pK}_a\), calculated as the negative logarithm of \(K_a\), is crucial in the Henderson-Hasselbalch equation. By knowing \(\text{pK}_a\), we can better understand the range over which an acid can effectively buffer, as well as when and how a color change in pH indicators will occur. This allows for precise adjustments and measurements in various chemical applications.