CH3COOH>> H+ + CH3COO-
k=(h+)(CH3COOH-)/.83
k=2H/.83
solve for H
I get about 1.7E-5*.83/2=.70E-5 about
the pH=-logH=about 5.1
Use your table for Ka, and you calculator to check, this is estimated.
I need help please. The question is vinegar, or a dilute of CH3COOH (acetic acid), should have a molarity of around 0.83 M. Using the Ka for acetic acid, what should the pH of vinegar be??
Thank you!!
2 answers
Given 0.84M HOAc and Ka =1.8x10^5
For a weak monoprotic acid the [H] can be calculated from [H] = (Ka[Acid)^1/2
= [1.75x10^-5(.83)]^1/2
= 3.8x10^-3M(H)
pH(vinegar) = -log[H] = -log(3.8E-3)
pH = 2.42
For a weak monoprotic acid the [H] can be calculated from [H] = (Ka[Acid)^1/2
= [1.75x10^-5(.83)]^1/2
= 3.8x10^-3M(H)
pH(vinegar) = -log[H] = -log(3.8E-3)
pH = 2.42