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!!

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.

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

To find the pH of vinegar, we need to use the information provided about the concentration of acetic acid and the Ka value. The Ka value represents the acid dissociation constant of acetic acid.

To solve this problem, we will use the equation for the acid dissociation constant:

Ka = [H+][CH3COO-] / [CH3COOH]

Given that the molarity of acetic acid (CH3COOH) is 0.83 M, we can assume that the concentration of acetate ions (CH3COO-) is also 0.83 M since the compound is a 1:1 acid-base reaction.

Therefore, we can rewrite the equation as:

Ka = [H+] * 0.83 M / 0.83 M

Simplifying the equation gives us:

Ka = [H+]

Now we can solve for [H+], which represents the concentration of hydrogen ions.

Since the Ka value for acetic acid is 1.8 x 10^-5 (given in most chemistry textbooks or resources), we can substitute the Ka value into the equation:

1.8 x 10^-5 = [H+]

Taking the negative logarithm (base 10) of both sides, we find the pH:

pH = -log [H+]

Thus, we have:

pH = -log (1.8 x 10^-5)

Evaluating this expression gives the pH of vinegar.

Please note that the actual pH value of vinegar can vary depending on factors like impurities, different concentrations, and temperature. However, using this method will give you a reasonable approximation.