Question

The indicator dinitrophenol has a Ka of 1.1 × 10⁻⁴. In a 1.0 × 10⁻⁴-M solution, it is colorless in acid and yellow in base. Calculate the pH range over which it goes from 10% ionized (colorless) to 90% ionized (yellow).

a) pH 3.5 to 5.0
b) pH 4.0 to 6.0
c) pH 5.5 to 7.0
d) pH 6.0 to 8.0

Answer 1

The pH range over which dinitrophenol goes from 10% ionized to 90% ionized is from pH 3 to pH 5, calculated using the Henderson-Hasselbalch equation with the provided acid dissociation constant (Ka). The correct option is A.

Step-by-step explanation:

To calculate the pH range over which dinitrophenol goes from 10% ionized (colorless) to 90% ionized (yellow), we need to use the Henderson-Hasselbalch equation.

The Ka for dinitrophenol is given as 1.1 × 10⁻⁴. At 10% ionization, the ratio [In⁻]/[HIn] is 0.1, and at 90% ionization, this ratio is 10. By inserting these ratios and the given Ka into the equation, we can solve for the pH values at these percentages.

The Henderson-Hasselbalch equation is: pH = pKa + log([In⁻]/[HIn]). For 10% ionization, the equation becomes pH = -log(1.1 × 10⁻⁴) + log(0.1) = 4 - log(10) = 4 - 1 = 3. For 90% ionization, the equation is pH = -log(1.1 × 10⁻⁴) + log(10) = 4 + 1 = 5. Therefore, the pH range is from 3 to 5 for the transition from 10% ionized to 90% ionized.