The $K_a$ of $CH_3COOH$ is $1.8 \times 10^{-5}$. How many grams of $CH_3COONa$ are required in $0.1 \ M \ CH_3COOH$ to form a solution having $pH = 4.0$? (Molecular mass of $CH_3COONa = 82 \ g \ mol^{-1}$)

(A) Using the Henderson-Hasselbalch equation for an acidic buffer: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$.
Given $pH = 4.0$ and $K_a = 1.8 \times 10^{-5}$,we find $pK_a = -\log(1.8 \times 10^{-5}) \approx 4.74$.
Substituting the values: $4.0 = 4.74 + \log \frac{[CH_3COONa]}{0.1}$.
$-0.74 = \log \frac{[CH_3COONa]}{0.1}$.
$\frac{[CH_3COONa]}{0.1} = 10^{-0.74} \approx 0.182$.
$[CH_3COONa] = 0.182 \times 0.1 = 0.0182 \ M$.
Mass of $CH_3COONa = \text{Molarity} \times \text{Volume} \times \text{Molar Mass}$. Assuming $1 \ L$ solution: $0.0182 \ mol \times 82 \ g \ mol^{-1} = 1.4924 \ g$.