Final answer:
To calculate the pH of a 0.045 M solution of weak acid Hn with Ka=2.55 × 10⁻⁴, we use an ice table to find the concentration of H⁺ ions, then use this value to calculate the pH with the equation pH = -log[H⁺].
Step-by-step explanation:
To calculate the pH of a 0.045 M Hn weak acid solution with a Ka of 2.55 × 10⁻⁴, we need to set up an ice table (initial, change, equilibrium) for the dissociation of the acid:
Hn → H⁺ + n⁻
We plug these into the expression for the acid dissociation constant:
Ka = [H⁺][n⁻] / [Hn]
2.55 × 10⁻⁴ = (x)(x) / (0.045-x)
Since Ka is relatively small, we can assume that x is much smaller than 0.045 M and simplify the expression to:
2.55 × 10⁻⁴ = x² / 0.045
Solving for x, we find x = √(2.55 × 10⁻⁴ × 0.045), which gives us the concentration of H⁺ ions at equilibrium.
Using this value of x, we can calculate the pH:
pH = -log[H⁺]
Finally, the calculated x is the hydrogen-ion concentration, and substituting it into the pH equation gives us the pH.
For example, if x is found to be 3.0 × 10⁻ M, the pH is -log(3.0 × 10⁻) = 2.52.