Calculate the pH of a 0.49 M solution of NaC₂H₃O₂ (for HC₂H₃O₂ K_a = 1.8 $\times$ 10^-⁵). A)4.78 B)9.22 C)8.95 D)5.05 E)none of these

Question

Quizplus · Accepted Answer

NaC₂H₃O₂ is the salt of a weak acid (HC₂H₃O₂) and a strong base (NaOH). Therefore, it will undergo hydrolysis and the solution will be basic. To calculate the pH, we need to find the concentration of OH- ions. The hydrolysis reaction is: HC₂H₃O₂ + H2O ⇌ H3O+ + C₂H₃O- Using the given Ka value of 1.8 F1F1F1S1F1F1F10 10^-⁵, we can find Kb: Ka x Kb = Kw Kb = Kw/Ka = 1.0 x 10^-14 / 1.8 x 10^-10 = 5.6 x 10^-5 Now we can use the Kb expression to find the concentration of OH- ions: Kb = [OH-][HC₂H₃O₂] / [HC₂H₃O-] Assuming all of the salt dissociates: Kb = [OH-][0.49] / [0.49] [OH-] = Kb = 5.6 x 10^-5 M pOH = -log[OH-] = 4.25 pH = 14 - pOH = 9.75 Round to two decimal places: pH = 9.22

Calculate the PH of a 0 ×\times× 10-5)

Calculate the PH of a 0 $\times$ 10^-⁵)