Question

A certain acid-base indicator is red in acidic solution and blue in basic solution. At \(\mathrm{pH}=5\), \(75 \%\) of the indicator is present in the solution in its blue form. Calculate the \(\mathrm{pH}\) at which the indicator shows \(90 \%\) red form? (Given \(10^{-4.523}=3 \times 10^{-5}\) ) (1) \(3.56\) (2) \(5.47\) (3) \(2.5\) (4) \(7.4\)

Short answer

The pH at which the indicator shows 90% red form is approximately 3.56.

Step-by-step solution

Understand the problem

The problem provides the percentages of the acid-base indicator present in two different forms at different pH levels. Given that at \(\text{pH} = 5\), the indicator is 75% in its blue form, we need to find the pH at which it shows 90% red form.

Define the relationship between forms and pH

Use the Henderson-Hasselbalch equation: \(\text{pH} = \text{pKa} + \text{log} \frac{[\text{A}^-]}{[\text{HA}]}\), where \([\text{A}^-]\) is the concentration of the ionized (blue) form and \([\text{HA}]\) is the concentration of the non-ionized (red) form.

Calculate the ratio at \(\text{pH}=5\)

Given at \(\text{pH}=5\), 75% of the indicator is in the blue form, let \([A^-] = 0.75 \) and \([HA] = 0.25\). Substitute these values into the ratio: \(\frac{[\text{A}^-]}{[\text{HA}]} = \frac{0.75}{0.25} = 3\)

Determine \(\text{pKa}\)

Using the Henderson-Hasselbalch equation, substitute \(\text{pH} = 5\), \(\frac{[\text{A}^-]}{[\text{HA}]} = 3\): \(\text{pKa} = 5 - \text{log}(3)\). We know that \(\text{log}(3) \approx 0.477\), thus \(\text{pKa} = 5 - 0.477 = 4.523\)

Calculate the ratio for 90% red form

If the indicator is 90% in its red form, then \([HA] = 0.90\) and \([A^-] = 0.10\), giving \(\frac{[\text{A}^-]}{[\text{HA}]} = \frac{0.10}{0.90} = \frac{1}{9}\)

Find the new pH at 90% red form

Substitute into the Henderson-Hasselbalch equation: \(\text{pH} = 4.523 + \text{log}\frac{1}{9}\). The \(\text{log}\frac{1}{9} = -\text{log}(9) = -\text{log}(3^2) = -2 \text{log}(3) \approx -2 \times 0.477 = -0.954\). Therefore, \(\text{pH} = 4.523 - 0.954 = 3.569\)

Compare with provided answers

From the calculations, the closest value to \(3.569\) among the options provided is (1) \(3.56\).

Key concepts

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a fundamental tool in understanding acid-base equilibrium. It relates the pH of a solution to the ratio of the concentration of its ionized and non-ionized forms. The equation is given by:

\( \text{pH} = \text{pKa} + \text{log} \frac{[\text{A}^-]}{[\text{HA}]} \)

In the formula, \( \text{pKa} \) is the dissociation constant of the acid, \( [\text{A}^-] \) is the concentration of the conjugate base (ionized form), and \( [\text{HA}] \) is the concentration of the acid (non-ionized form). This equation helps us determine the pH of a buffer solution:

• pKa: It is a measure of the strength of the acid. Smaller pKa values indicate stronger acids.

This equation is especially useful for calculating the pH in systems where weak acids or bases and their conjugates are present in a solution.

pH Calculation

pH is a measure of the acidity or basicity of a solution. It is calculated using the negative base-10 logarithm of the concentration of hydrogen ions (H\textsuperscript{+}) in the solution:

\( \text{pH} = -\text{log} [\text{H}^+] \)

To understand the pH calculation in the context of the original exercise, consider the given information:

1. At pH 5, 75% of the indicator is in its blue form, meaning that 25% is in its red form.
2. This ratio helps us calculate the pKa of the indicator using the Henderson-Hasselbalch equation.
3. Using pKa, further pH calculations can be performed for different concentrations of ionized and non-ionized forms, such as the situation where 90% of the indicator is in its red form.

The calculation steps involve:

• Finding the ratio of concentrations based on given percentages.
• Applying the ratios in the Henderson-Hasselbalch equation to find pKa.
• Using the obtained pKa for further pH calculations at different indicator proportions.

These steps are interconnected, and a solid grasp of each is essential in understanding pH calculations with indicators.

Acid-Base Equilibrium

Acid-base equilibrium refers to the state of balance between the concentrations of acids and their corresponding bases. In the exercise, we are dealing with an acid-base indicator that changes color based on the pH of the solution. This change occurs due to a shift in the equilibrium state of the indicator's ionized and non-ionized forms.

An acid-base indicator is a valuable tool in determining the pH of a solution visually. Here are the basic principles:

• Indicators have different colors in their protonated (acidic) and deprotonated (basic) forms.
• The pH at which the indicator changes color is related to its pKa value.

In the practical application of acid-base equilibrium to the problem:

• We used the known percentages of the indicator's colors at given pH values to find the ratio of ionized (blue) to non-ionized (red) forms.
• These ratios, when placed into the Henderson-Hasselbalch equation, provide insight into the acid-base behavior of the indicator.

Understanding acid-base equilibrium helps us predict and interpret the behavior of indicators and the resulting pH changes.