Question

A buffer is made by mixing 60.0 mL of 0.5 M acetic acid and 40.0 mL of 0.50 M sodium acetate. What is the pH of the buffer solution?

Answer 1

pH = 4.5809

Step-by-step explanation:

balanced reaction:

• CH3COONa → CH3COO- + Na+
• CH3COOH ↔ CH3COO- + H3O+
• 2H2O ↔ H3O+ + OH-

∴ Ka = ([CH3COO-]*[H3O+]) / [CH3COOH] = 1.75 E-5

∴ C CH3COONa = 0.50 M

∴ V CH3COONa = 0.04 L

∴ C CH3COOH = 0.50 M

∴ V CH3COOH = 0.06 L

mass balance:

⇒ C CH3COONa + C CH3COOH = [CH3COO-] + [CH3COOH] = 1 M.....(1)

charge balance:

⇒ [H3O+] + [Na+] = [CH3COO-] + [OH-]

∴ [Na] = 0.50 M

∴ [OH-]: is neglected, come frome to water.

⇒ [H3O+] + 0.50 = [CH3COO-]..............(2)

(2) in (1):

⇒ 1 M = ([H3O+] + 0.50) + [CH3COOH]

⇒ [CH3COOH] = 0.50 - [H3O+]

replacing in Ka:

⇒ Ka = 1.75 E-5 = ([H3O+]*([H3O+] + 0.50)) / (0.50 - [H3O+])

⇒ 1.75 E-5 = ([H3O+]² + 0.50[H3O+]) / (0.50 - [H3O+])

⇒ 8.75 E-6 - 1.75 E-5[H3O+] = [H3O+]² + 0.50[H3O+]

⇒ [H3O+]² + 0.50[H3O+] - 8.75 E-6 = 0

⇒ [H3O+] = 2.625 E-5 M

• pH = - Log[H3O+]

⇒ pH = - Log(2.625 E-5)

⇒ pH = 4.5809