what's the pH of
the acid HCl with concentration of .15?
the acid HC2H3O2 with concentration of .15?
I figured out the pH of HCl, but I'm still not sure how to find the other one...
What is the dissociation constant for acetic acid?
K= [H+][ace-]/.15
pH= -log sqrt (.15 k) check that.
acetic acid being weak is ionized in less quantity in the solution.its dissociation constant is taken to be 0.0000176 By the given problem [H+]=root over(dissociation constant*given molarity) finally the pH obtained is 2.789198037
To find the pH of the acid HC2H3O2 with a concentration of 0.15, you need to know the dissociation constant (K) of acetic acid. The dissociation constant represents the degree of ionization of the acid in water.
First, let's find the dissociation constant for acetic acid. Acetic acid is represented by the chemical formula HC2H3O2. In water, it dissociates into hydrogen ions (H+) and acetate ions (ace-).
The dissociation of acetic acid can be represented as follows:
HC2H3O2 ⇌ H+ + ace-
Now, let's denote the concentration of acetic acid as [HC2H3O2]. According to the dissociation equation, the concentration of hydrogen ions ([H+]) and acetate ions ([ace-]) will also be [HC2H3O2] since the acid dissociates in a 1:1 ratio.
The expression for the dissociation constant (K) is given by:
K = [H+][ace-]/[HC2H3O2]
Since you have the concentration of acetic acid (0.15), you can substitute it into the equation:
K = [H+][ace-]/0.15
Now, to find the pH of an acid, you can use the equation:
pH = -log[H+]
Since the concentration of hydrogen ions is equal to the concentration of acetic acid (0.15), you can write it as:
[H+] = 0.15
Using the dissociation constant, you can write the pH equation as:
pH = -log√(0.15 * K)
Substituting the value of K you obtained earlier, you can find the pH of the acid HC2H3O2 with a concentration of 0.15.