Question

What is the pH of a solution that is 0.500 M in acetic acid and 1.00 M in CH3COONa? Ka = 1.75*10-5 Group of answer choices 4.47 5.06 4.77 0.3 Flag this Question

Answer 1

5.06

Step-by-step explanation:

Considering the Henderson- Hasselbalch equation for the calculation of the pH of the buffer solution as:

pH=pKa+log[base]/[acid]

Where Ka is the dissociation constant of the acid.

Given that the acid dissociation constant = 1.75×10⁻⁵

pKa = - log (Ka) = - log (1.75×10⁻⁵) = 4.76

Given concentration of acid = [acid] = 0.500 M

[Base] = 1.00 M

`pH=4.76+\log (1.00)/(0.500) =5.06`