Question

For the following dissociation of acetic acid, K_a = 1.760 x 10^-5:

CH₃COOH + H₂O ? CH₃COO^- + H₃O⁺

Sodium acetate completely dissociates in solution according to the following reaction:

NaCH₃COO ? Na⁺ + CH₃COO^-

A solution was prepared which contains CH₃COOH at a pre-equilibrium concentration of 0.05782 M and NaCH₃COO at a pre-equilibrium concentration of 0.04991 M.
I know all answers, just do not know how to find them.
What is the equilibrium concentration of H₃O⁺? (Answer: 0.00002037)

What is the molar equilibrium concentration of CH₃COO^-? (Answer: 0.04993)

What is the molar equilibrium concentration of CH₃COOH? (Answer: 0.05780)

What is the pH of this solution? (Answer: 4.69)

Answer 1

H_3 c - cooH + H_2 0 leftrightarrow H_3 c - coo^(-) + H_3 o^(+) Na cH_3coo + rightarrow H_3 c - coo^(-) + Na^(+) common ian effect H_3 c - cooH + H_2 o() rightarrow H_3 c - coo^(-) + H_3 o^(+) Initial 0.05782 M 0.04991 M - change - x + x + x equilibrium 0.05782 - x 0.04991 + x x K_a = [cH_3 coo^(-)] (H_3 o^+]/[cH_3 cooH] 1.76 times 10^-5 = (0.04991 + x) . x/(0.05782 - x) . x = 2.037 times 10^-5 at equilibrium, [H_3 o^+] = x = 2.037 times ^-5 m . therefore [(cH_3 coo^(-)] = 0.04991 + 2.037 times 10^-5 = 0.04993 M [cH_3 cooH] = 0.05782 - 2.037 times 10^-5 = 0.05780 M P^H = -log (H_3 o^+) = -log (2.037 times 10^-5) P^H = 4.69