At what pH is acetic acid 30% ionized?
Call acetic acid HAc
...............HAc---> H^+ + Ac^-
initial.........x......0......0
change.........-0.3x..+0.3x...+0.3x
equil........x-0.3x..... 0.3x.....0.3x
Ka = (H^+)(Ac^-)/(HAc)
Substitute equil conditions into Ka expression and solve for (H^+), then convert to pH.
I don't get how you put 30% ionization into te problem. So far I have: pH = 4.76 + log[A^-]/[HAc^-]
To determine the pH at which acetic acid (CH3COOH) is 30% ionized, we need to consider its acid dissociation constant, Ka, which describes the extent of its ionization in water.
The dissociation of acetic acid can be represented by the equation:
CH3COOH ⇌ CH3COO- + H+
The Ka expression for acetic acid is:
Ka = [CH3COO-][H+] / [CH3COOH]
Since acetic acid is a weak acid, we can assume that the concentration of CH3COOH that ionizes is very small compared to the concentration of the initial acetic acid, so we can simplify the expression to:
Ka ≈ [CH3COO-][H+]
To determine the pH at which acetic acid is 30% ionized, we need to find the concentration of [H+] when [CH3COO-] is 30% of the initial concentration of acetic acid.
Let's assume the initial concentration of acetic acid is C, then the concentration of CH3COO- at 30% ionization is 0.3C. Since acetic acid dissociates in a 1:1 ratio, the concentration of [H+] will also be 0.3C.
Using the equation for pH, pH = -log[H+], we can calculate the pH.
pH = -log(0.3C)
Note: The specific value of C is not provided in the question, so we cannot determine the exact pH without knowing the initial concentration of acetic acid.