Question

The pH of an aqueous buffer prepared using CH3COOH and CH3COO Na⁺ is 4.80.
The quantity \(\frac{[CH_3COO^-]-[CH_3COOH]}{[CH_3COOH]}\) is _____. (round off to three decimal places)
[Given: pK_a of CH₃COOH in water is 4.75]

Answer 1

Correct Answer: 0.121

To solve for \(\frac{[CH_3COO^-]-[CH_3COOH]}{[CH_3COOH]}\), we use the Henderson-Hasselbalch equation:

\[\text{pH} = \text{p}K_a + \log\left(\frac{[CH_3COO^-]}{[CH_3COOH]}\right)\]

Given:
pH = 4.80
pK\(_a\) of CH\(_3\)COOH = 4.75

Substitute the values:
4.80 = 4.75 + \log\left(\frac{[CH_3COO^-]}{[CH_3COOH]}\right)

Solve for \(\log\left(\frac{[CH_3COO^-]}{[CH_3COOH]}\right)\):
\[\log\left(\frac{[CH_3COO^-]}{[CH_3COOH]}\right) = 4.80 - 4.75 = 0.05\]

Convert the logarithm to a ratio:
\[\frac{[CH_3COO^-]}{[CH_3COOH]} = 10^{0.05}\]
Calculate \(10^{0.05}\):
\(10^{0.05} \approx 1.122\)

Thus, \(\frac{[CH_3COO^-]-[CH_3COOH]}{[CH_3COOH]}\) can be determined:

\(\frac{[CH_3COO^-]-[CH_3COOH]}{[CH_3COOH]} = \left(\frac{[CH_3COO^-]}{[CH_3COOH]}-1\right)\)
Substitute the calculated ratio:
\(\frac{[CH_3COO^-]-[CH_3COOH]}{[CH_3COOH]} = 1.122 - 1 = 0.122\)

Round to three decimal places: **0.122**