What is the hydronium ion concentration of a 0.100 M acetic acid solution with Ka = 1.8 × 10-5? The equation for the dissociation of acetic acid is: CH3CO2H(aq) + H2O(l) ⇌ H3O+(aq) + CH3CO2-(aq) What is the hydronium ion concentration of a 0.100 M acetic acid solution with Ka = 1.8 × 10-5? The equation for the dissociation of acetic acid is: CH3CO2H(aq) + H2O(l) ⇌ H3O+(aq) + CH3CO2-(aq) 1.3 × 10-2 M 4.2 × 10-2 M 1.3 × 10-3 M 4.2 × 10-3 M

1.3×10^-3 M

Explanation:

Step 1:

Data obtained from the question:

Equilibrium constant (Ka) = 1.8×10^-5

Concentration of acetic acid, [CH3COOH] = 0.100 M

Concentration of hydronium ion, [H3O+] =..?

Step 2:

The balanced equation for the reaction.

CH3CO2H(aq) + H2O(l) ⇌ H3O+(aq) + CH3CO2-(aq)

Step 3:

Determination of concentration of hydronium ion, [H3O+].

This can be obtained as follow:

Ka = [H3O+] [CH3CO2-] / [CH3CO2H]

Initial concentration:

[CH3COOH] = 0.100 M

[H3O+] = 0

[CH3CO2-] = 0

During reaction

[CH3COOH] = – y

[H3O+] = +y

[CH3CO2-] = +y

Equilibrium:

[CH3COOH] = 0.1 – y

[H3O+] = y

[CH3CO2-] = y

Ka = [H3O+] [CH3CO2-] / [CH3CO2H]

1.8×10^-5 = y × y / 0.1

Cross multiply

y^2 = 1.8×10^-5 x 0.1

Take the square root of both side

y = √(1.8×10^-5 x 0.1)

y = 1.3×10^-3 M

[H3O+] = y = 1.3×10^-3 M

Therefore, the concentration of the hydronium ion, [H3O+] is 1.3×10^-3 M