QUESTION: Calculate the pH of a 0.05 M solution of ascorbic acid (Ka1 = 7.9x10–5; Ka2 = 1.6x10–12). SHOW YOUR WORK STEP BY STEP
for weak acid Ka2 plays essentially no part in the H3O^+
conc.
HA + H2O ⇋ H3O^+ + A^-
Ka = [H3O^+][A-]/[HA] = 7.9×10^-5
let [H3O^+] = [A-] = x; [HA] = 0.05 - x ≈ 0.05
(x)(x)/0.05 = 7.9×10^-5
x = √(3.9×10^-6) = 1.97×10^-3
pH = -log[H3O^+] = 2.70
QUESTION: Calculate the pH of a 0.05 M solution of ascorbic acid (Ka1 = 7.9x10–5; Ka2...
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