Question

What is the pH of a 0.2 M acetate buffer (pKa = 4.77) that contains twice as much acid as conjugate base?

Answer 1

Answer:

The pH of the buffer is 5.07

Explanation:

pKa = 4.77 so Ka = 10^(-4.77), and the formula of Ka is as below

$K_{a} = \frac{[CH_{3}COOH] . [H^{+}]}{[CH_{3}COO^{-}]} = 10^{-4.77}$

The solution contains twice as much acid as conjugate base, so the concentration of acid form (CH₃COOH) is double the base form (CH₃COO⁻).

Meanwhile, the concentration of the solution is 0.2M so the total of (CH₃COOH) and (CH₃COO⁻) in 1 L of the solution is also 0.2 mol.

Then we get the equation as below, which a is the concentration of CH₃COOH and b is the concentration of CH₃COO⁻

$\left \{ {{a=2b} \atop {a+b=0.2}} \right.$

Solve this one, we get a = 4/30, b=2/30, use these number, add to the Ka formula, we can find [H⁺] = 10^(-4.77) x b / a = 0.5 x 10^-4.77 = 10^(-5.07)

Hence pH = 5.07