4. Write the expression for Ka for the ionization of acetic acid in water.
Ka= [H3O^+][CH3COO^-] / [CH3COOH]
Write the expression for Ka for the ionization of acetic acid in water.
CH3COOH ==> CH3COO^- + H^+
Ka = (H^+)(CH3COO^-)/(CH3COOH)
When I calculate the percent difference between your value for Ka and the accepted value, what could be two sources of error that might account for any differences?
I'm sorry but I don't understand the question. I didn't give you a value for Ka so how can there be a difference? I gave you the Ka expression but not a value for Ka.
To write the expression for Ka, we need to understand that Ka represents the acid dissociation constant. It measures the extent to which an acid will ionize in water.
The ionization of acetic acid (CH3COOH) in water can be represented by the following chemical equation:
CH3COOH + H2O ⇌ CH3COO- + H3O+
In this equation, acetic acid (CH3COOH) donates a proton (H+) to water (H2O) and forms the acetate ion (CH3COO-) and hydronium ion (H3O+).
Now, let's write the expression for Ka:
Ka = [CH3COO-][H3O+] / [CH3COOH]
In this expression, [CH3COO-] represents the concentration of the acetate ion, [H3O+] represents the concentration of the hydronium ion, and [CH3COOH] represents the concentration of acetic acid.
Please note that the expression only includes aqueous species because Ka is specifically defined for reactions occurring in water.