Question

Suppose there is 1.00 L of an aqueous buffer containing 60.0 mmol of acetic acid ( p K a = 4.76 ) and 40.0 mmol of acetate. Calculate the pH of this buffer.

Answer 1

Answer:

pH = 4.584

Explanation:

• CH3COOH ↔ CH3COO-  +  H3O+

for buffer solution, Henderson-Hauselbach equation:

• pH = pKa + log [A-]/[HA]

∴ HA: CH3COOH

∴ pKa = 4.76

⇒ [CH3COOH] = ((60.0 mmol)×(mol/1000 mmol))/(1.00 L) = 0.06 M

⇒ [CH3COO-] = ((40.0 mmol)×(mol/1000 mmol)/(1.00 L) = 0.04 M

⇒ pH = 4.76 + log (0.04/0.06)

⇒ pH = 4.76 + ( - 0.176 )

⇒ pH = 4.584