What is the [H3O] concentration in a 1.0M CH3COONa solution? (The Ka for CH3COONa is 2.0*10-5.)
To find the [H3O+] concentration in a solution of CH3COONa, we need to understand the concept of a salt dissolving in water and the relationship between the acid and its conjugate base.
CH3COONa is the salt of acetic acid (CH3COOH) and its conjugate base acetate ion (CH3COO-). When CH3COONa dissolves in water, it dissociates into its constituent ions. In this case, it dissociates into CH3COO- and Na+ ions.
The dissociation of acetic acid (CH3COOH) can be represented by the following equation:
CH3COOH(aq) ⇌ H3O+(aq) + CH3COO-(aq)
The dissociation equation indicates that one molecule of acetic acid produces one hydronium ion (H3O+) and one acetate ion (CH3COO-).
Now, let's consider the equilibrium expression for this dissociation:
Ka = [H3O+][CH3COO-] / [CH3COOH]
Given that the Ka for CH3COONa is 2.0 × 10^-5, we can use this equilibrium expression to determine the [H3O+] concentration.
Since we are given the concentration of CH3COONa as 1.0M, we can assume that the initial concentration of CH3COOH is negligible compared to the concentration of CH3COONa. Therefore, we can simplify the equilibrium expression to:
Ka ≈ [H3O+][CH3COO-] / [1.0M]
Rearranging the equation, we get:
[H3O+] ≈ Ka * [CH3COO-]
Now, we can plug in the given value of Ka and the concentration of CH3COONa into the equation:
[H3O+] ≈ (2.0 × 10^-5) * (1.0M)
Calculating this expression, we find that the approximate [H3O+] concentration in a 1.0M CH3COONa solution is 2.0 × 10^-5 M.
The pH is determined by the hydrolysis of the salt. Let Ac = acetate; HAc = acetic acid
.........Ac^- + HOH ==> HAc +OH^-
initial...1M.............0....0
change....-x.............x.....x
equil....1-x.............x.....x
Kb for Ac^- = (Kw/Ka for HAc) = (HAc)(OH^-)/(Ac^-)
Substitute from the ICE chart and solve for x, convert to (H^+).