What is Ph of the following solution with its known Ka of 1.8x10^-5?

Question

CH3COOH (100mL of 1.00M) + OH(100mL of 1.0M)<-->H3O+CH3COONa(aq)

Answer 1

The correct reaction equation is:

CH3COOH (100mL of 1.00M) + NaOH (100mL of 1.0M) → H2O + CH3COONa

We can see the the same number of moles of acetic acid and sodium hydroxide were mixed. The volume of the resultant sodium acetate solution is 200 mL, but the number of moles is the same. Therefore, in reality we are trying to find the pH of a 0.5 M solution of sodium acetate. This is done assuming &quot;hydrolysis&quot; of acetate:

H2O + CH3COO- ↔ CH3COOH + OH-

The equilibrium constant for this is related to Ka by

KaKb = Kw, so the equilibrium constant Kb is:

(1x10^-14) / (1.8x10^-5) = 5.56x10^-10.

We can take the shortcut to calculate [OH-]:

[OH-] ≈ √(Kb*C = (5.56x10^-10) (0.5 M) = 1.67x10^-5 M

pOH = -log(1.67x10^-5 M) = 4.8.

pH = 14-pOH = 14 - 4.8 = 9.2, the answer.