Final answer:
The pH of a 0.1 M solution of ascorbic acid is approximately 2.05.
Step-by-step explanation:
The pH of a solution is a measure of its acidity or alkalinity. In the case of ascorbic acid (vitamin C), it can act as a weak acid when dissolved in water.
To find the pH of a 0.1 M solution of ascorbic acid, we need to determine its equilibrium concentration of hydrogen ions ([H+]). Since ascorbic acid is a weak acid, we can use the Ka expression to find [H+] using the concentration of ascorbic acid.
The Ka expression for ascorbic acid is:
Ka = [H+][C₆H₆O₆] / [C₆H₈O₆]
Given that we have a 0.1 M solution of ascorbic acid, we can assume that the initial concentration of ascorbic acid is also 0.1 M. Let x be the equilibrium concentration of [H+]. The equilibrium expression becomes:
Ka = x² / (0.1 - x)
Since ascorbic acid is a weak acid, we can assume that x is small compared to 0.1. This means that 0.1 - x ≈ 0.1. We can simplify the expression:
Ka ≈ x² / 0.1
To find x, we use the value of the Ka for ascorbic acid, which is 8.0 x 10^-5. Rearranging the equation, we get:
x² = Ka * 0.1
x = √(Ka * 0.1)
Substituting the value of Ka, we find:
x ≈ √(8.0 x 10⁻⁵ * 0.1)
x ≈ 8.94 x 10⁻³
Since we assumed that x is small compared to 0.1, we can approximate [H+] as the same value as x.
Therefore, the [H+] is approximately 8.94 x 10⁻³ M. To find the pH, we use the equation:
pH = -log[H+]
pH = -log(8.94 x 10⁻³)
pH ≈ 2.05