Answer:
% α = 1.7243 %
Step-by-step explanation:
- % ionization (α) = α*100
- HCOOH ↔ H+ + COOH-
∴ Ka = 1.78 E-4 = [H+][COOH-]/[HCOOH]
∴ α = [COOH-] / [COOH-] + [HCOOH]
∴ C HCOOH = 0.580 M
mass balance:
⇒ C HCOOH = 0.580 = [HCOOH] + [COOH-]
charge balance:
⇒ [H+] = [COOH-]
Replacing:
⇒ Ka = 1.78 E-4 = [H+]²/(0.580 - [H+])
⇒ [H+]² + 1.78 E-4[H+] - 1.0324 E-4 = 0
⇒ {H+] = 0.01 M
⇒ [COOH-] = 0.01 M
⇒ [HCOOH] = 0.580 - 0.01 = 0.5699 M
⇒ α = (0.01)/(0.01 + 0.5699) = 0.0172
⇒ % α = 1.7243 %