Ka= (4e-3)^2/.9=1.777e-5 M
assuming both the concentrations you gave were at equilibrium
CH3COOH(aq) + H2O(l) ⇌ H3O^+(aq) + CH3COO^–(aq)
[H3O^+] = 4.0 × 10^–3 M
[CH3COOH] = 0.90 M
What is the Ka value for the reaction?
1 answer