Answer to Question #76906 in Chemistry for Dollar

Question

What is the pH of 0.1 mol/DM of ethanoic acid which has a Ka of 1.74 * 10^-6

Answer 1

pKa = -log(10)[Ka], pka=- log10 [ 1.74×10^-6] =5.75
Ka for acetic acid is [H+][CH3COO-]/[CH3COOH] where [] indicates the concentration.
pKa = log(10)[CH3COOH] - log(10)[H+] - log(10)[CH3COO-]
as [H+] = [CH3COO-]
pKa = log(10)[CH3COOH] - 2*log(10)[H+]
For acetic acid pKa = 5.75 and -log(10)[H+] = pH
5.75= log(10)(0.1) + 2*pH
=> 2*pH = 5.75 + 1
=> pH = 6.75/2
=> pH = 3.38
The required pH of 0.1 M solution of acetic acid is 3.38