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 [tex]pH = pK_a + log(\frac{[A^-]}{[HA]})[/tex]. Firstly, let's find the [tex]pK_a[/tex] value:
[tex]pK_a = -log(K_a) = -log(1.8\cdot 10^-5) = 4.74[/tex]
Now, for buffer A:
[tex]\frac{[A^-]}{[HA]}=\frac{[CH_3COO^-]}{[CH_3COOH]}=\frac{1}{10}[/tex]
This means:
[tex]pH = 4.74 + log(\frac{1}{10}) = 3.74[/tex]
Similarly, for buffer B:
[tex]\frac{[A^-]}{[HA]}=\frac{[CH_3COO^-]}{[CH_3COOH]}=\frac{10}{1}[/tex]
This means:
[tex]pH = 4.74 + log(\frac{10}{1}) = 5.74[/tex]
Similarly, for buffer C:
[tex]\frac{[A^-]}{[HA]}=\frac{[CH_3COO^-]}{[CH_3COOH]}=\frac{1}{1}[/tex]
This means:
[tex]pH = 4.74 + log(\frac{1}{1}) = 4.74[/tex]
