Question

The \(pK_a\) of acetic acid is \(4.74\). The concentration of \( \mathrm{CH_3COOH} \) is \(0.01\, M\). The pH of \( \mathrm{CH_3COOH} \) is _ _ _ _ _ .

Hint:

For weak acids: \(\text{pH} = \frac{1}{2}(pK_a - \log C)\) — memorize this shortcut!

Answer 1

Correct Answer: 3.37

Concept: For a weak acid: \[ \text{pH} = \frac{1}{2}(pK_a - \log C) \]

Step 1: Given: \[ pK_a = 4.74, \quad C = 0.01 \, M = 10^{-2} \, M \]

Step 2: \[ \log C = \log(10^{-2}) = -2 \] \[ \text{pH} = \frac{1}{2}[4.74 - (-2)] = \frac{1}{2}(4.74 + 2) \] \[ \text{pH} = \frac{1}{2}(6.74) = 3.37 \]