Question

The Ka of acetic acid is 1.76 × 10-5. The pH of a buffer prepared by combining 50.0 mL of 1.00 M potassium acetate and 50.0 mL of 1.00 M acetic acid is __________.

Answer 1

Answer:

pH = 4.75

Explanation:

Given:

Ka(acetic acid, CH3COOH) = 1.76*10^-5

Volume of CH3COOK = 50.0 ml

Molarity of CH3COOK = 1.00 M

Volume of CH3COOH = 50.0 ml

Molarity of CH3COOH = 1.00 M

To determine:

pH of the above buffer solution

Calculation:

The pH of a buffer is related to the concentrations of the conjugate base and acid via Henderson - Hasselbalch equation:

$pH = pKa + log\frac{[A-]}{[HA]}$

where pKa = -logKa

{A-] = conjugate base

[HA] = acid

For the CH3COOH/CH3COOK buffer:

$pH = pKa + log\frac{[CH3COOK]}{[CH3COOH]}$

here, [CH3COOK] = [CH3COOH], therefore,

$pH = pKa + log\(1 = pKa$

$pH = -logKa = -log(1.76*10^{-5} )=4.75$