Question

What will be the resulting pH if the solution initially contains 0.010 M of CH3COOH and 0.0050 M of NaCH3COOH? (please put ICE table)

Answer 1

CH3COOH + H2O ---> CH3COO- + H3O+

Initial 0.010 ---> 0.005 + 0

Change -x ---> +x +x

Equilibrium 0.010-x ---> 0.005+x +x

$K_a = \dfrac{[CH_3COO][H_3O^+]}{[CH_3COOH]}$

$[H_3O] = \dfrac{[CH_3COOH]×K_a}{[CHCOO-]}$

Ka

(CH3COOH) = 1.8 x 10-⁵

H3O = 0.005×1.8 x 10-⁵/0.010

H3O = 0.000009M = 9×10^{-6}

pH = -log[H3O+]

pH = -log[9×10-^{<¥

pH = 5.05

pH = 5.05.