Problem 67 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

For the indicator thymol blue, the value of \(\mathrm{pH}\) is \(2.0 \mathrm{when}\) half of the indicator is present in the unionized form. The percentage of the indicator in the unionized form in a solution of \(4.0\) \(\times 10^{-3} \mathrm{M}\) hydrogen ion concentration is (a) \(40 \%\) (b) \(28.6 \%\) (c) \(71.4 \%\) (d) \(60 \%\)

Step-by-Step Solution

Answer

The percentage of thymol blue in the unionized form is approximately 71.4%.

Step 1: Understand the problem and given data

We need to find the percentage of thymol blue in the unionized form when the hydrogen ion concentration is given as 4.0 × 10^{-3} M. It's given that the pH is 2.0 when half of the indicator is in the unionized form. The Henderson-Hasselbalch equation can be used to solve this problem.

Step 2: Convert hydrogen ion concentration to pH

Recall that pH is defined as the negative logarithm of hydrogen ion concentration. Therefore, the pH for 4.0 × 10^{-3} M hydrogen ion concentration can be calculated using pH = -log[H+].

Step 3: Calculate pH of the solution

Using the hydrogen ion concentration, calculate the pH: pH = -log(4.0 × 10^{-3}).

Step 4: Apply Henderson-Hasselbalch equation

Use the Henderson-Hasselbalch equation to find the ratio of the unionized form \(\frac{[In]}{[HIn^+]}\). The pH value at which 50% of the indicator is unionized is equal to the pKa value of the indicator. The equation is given by pH = pKa + log\(\frac{[In]}{[HIn^+]}\).

Step 5: Rearrange the equation

Rearrange the equation to solve for \(\frac{[In]}{[HIn^+]}\): \(\frac{[In]}{[HIn^+]}\) = 10^{(pH - pKa)}. Given that pKa = 2 to have 50% unionized, insert the calculated pH and solve for the ratio.

Step 6: Calculate percentage unionized

Finally, convert the ratio \(\frac{[In]}{[HIn^+]}\) to a percentage. The total amount is 100%, so unionized percentage is given by \(\frac{[In]}{[In]+[HIn^+]} \times 100\%\).

Key Concepts

pH Calculation

The pH of a solution is a measure of the acidity or alkalinity, based on the concentration of hydrogen ions present. The equation \(\text{pH} = -\log[H^+]\) quite literally provides a scale for measuring hydrogen ion concentration in the solution. \(\text{pH}\) stands for 'power of hydrogen'. The more hydrogen ions present, the lower the \(\text{pH}\), meaning the solution is more acidic. Conversely, fewer hydrogen ions yield a higher \(\text{pH}\) and a more alkaline solution.

For a quick calculation example, if a solution has a hydrogen ion concentration of \(4.0 \times 10^{-3} M\), the \(\text{pH}\) is calculated as follows:
\(\text{pH} = -\log(4.0 \times 10^{-3})\).
This formula is vital for chemists and biologists, as many biological processes are \(\text{pH}\)-sensitive and require precise \(\text{pH}\) control.

Unionized Form Percentage

The Henderson-Hasselbalch equation provides insight into the protonation state of a solution, which is crucial for understanding the behavior of acids and bases. Specifically, it helps determine the percentage of an acid in its unionized (non-protonated) form. As an approach for our exercise, wherein thymol blue acts as an acid-base indicator, this equation links \(\text{pH}\), \(\text{pKa}\) (the acid dissociation constant), and the ratio of concentrations of the unionized and ionized forms of the substance.

The equation is expressed as \(\text{pH} = \text{pKa} + \log\left(\frac{[In]}{[HIn^+]}\right)\), where \(\text{[In]}\) and \(\text{[HIn^+]}\) represent the concentrations of the unionized and ionized forms, respectively. To find the percentage in the unionized form, we rearrange to \(\text{[In]} = 10^{(\text{pH} - \text{pKa})} \times [HIn^+]\) and then calculate: \(\text{Unionized Percentage} = \frac{\text{[In]}}{\text{[In]} + \text{[HIn^+]}} \times 100\%\). Understanding this permits control over the protonation states, impacting solubility, absorption, and reactivity of substances in various fields such as pharmacology and chemistry.

Acid-Base Indicator

An acid-base indicator, like thymol blue in our solved problem, is a substance that changes color depending on the \(\text{pH}\) of the solution it is in, which reflects the protonation state of the indicator. The structure of the indicator changes as it gains or loses protons, which in turn affects the wavelengths of light it absorbs and thus its color. Indicators are chosen based on their \(\text{pKa}\) values which should be close to the \(\text{pH}\) at which the color change is desired.

For thymol blue, the given \(\text{pH}\) when 50% is unionized corresponds to its \(\text{pKa}\). This property makes it useful in titrations, where a sudden change in \(\text{pH}\) indicates an endpoint, and in various chemical analyses where an approximate \(\text{pH}\) range must be determined visually. The percentage of un-ionized thymol blue at a particular \(\text{pH}\), as calculated using the Henderson-Hasselbalch equation, helps to indicate the \(\text{pH}\) of the solution visually.