A buffer solution contains 0.0200 M acetic acid and 0.0200 M sodium acetate.What is the pH after 2.0 mmol of NaOH are added to 1.00 L of this buffer? pK_a = 4.75 for acetic acid.

Question

Quizplus · Accepted Answer

The addition of NaOH will consume the acetic acid and produce more acetate ions, shifting the equilibrium to the right. The new concentration of acetic acid can be calculated using the Henderson-Hasselbalch equation:```pH = pKa + log([A-]/[HA])```where [A-] is the concentration of acetate ions and [HA] is the concentration of acetic acid.The initial concentration of acetic acid is 0.0200 M, and the initial concentration of acetate ions is also 0.0200 M. After the addition of 2.0 mmol of NaOH, the concentration of acetic acid will decrease by 2.0 mmol/L, and the concentration of acetate ions will increase by 2.0 mmol/L. Therefore, the new concentrations of acetic acid and acetate ions will be:```[HA] = 0.0200 M - 2.0 mmol/L = 0.0180 M[A-] = 0.0200 M + 2.0 mmol/L = 0.0220 M```Substituting these values into the Henderson-Hasselbalch equation, we get:```pH = 4.75 + log(0.0220/0.0180) = 4.84```Therefore, the correct answer is D) 4.84.

A Buffer Solution Contains 0