Question

Acetic acid has a Ka of 1.8×10−5. Three acetic acid/acetate buffer solutions, A, B, and C, were made using varying concentrations:______[acetic acid] ten times greater than [acetate],[acetate] ten times greater than [acetic acid], and[acetate]=[acetic acid].Match each buffer to the expected pH.pH = 3.74 ; pH = 4.74 ; pH = 5.74

Answer 1

Answer:

A: pH = 3.74

B: pH = 5.74

C: pH = 4.74

Explanation:

According to the Henderson-Hasselbach equation for buffers, we know that $pH = pK_a + log(\frac{[A^-]}{[HA]})$. Firstly, let's find the $pK_a$ value:

$pK_a = -log(K_a) = -log(1.8\cdot 10^-5) = 4.74$

Now, for buffer A:

$\frac{[A^-]}{[HA]}=\frac{[CH_3COO^-]}{[CH_3COOH]}=\frac{1}{10}$

This means:

$pH = 4.74 + log(\frac{1}{10}) = 3.74$

Similarly, for buffer B:

$\frac{[A^-]}{[HA]}=\frac{[CH_3COO^-]}{[CH_3COOH]}=\frac{10}{1}$

This means:

$pH = 4.74 + log(\frac{10}{1}) = 5.74$

Similarly, for buffer C:

$\frac{[A^-]}{[HA]}=\frac{[CH_3COO^-]}{[CH_3COOH]}=\frac{1}{1}$

This means:

$pH = 4.74 + log(\frac{1}{1}) = 4.74$