aledechristine aledechristine
  • 25-04-2020
  • Chemistry
contestada

What is the percent ionization in a 0.580 M solution of formic acid ( HCOOH) (Ka=1.78*10^-4)?

Respuesta :

gbustamantegarcia054 gbustamantegarcia054
  • 26-04-2020

Answer:

% α = 1.7243 %

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 %

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