What will be the change in $pH$ by adding $0.1 \ M$ $CH_3COONa$ to $0.1 \ M$ $CH_3COOH$ at $298 \ K$ temperature? (Given: $pK_a$ of $CH_3COOH = 4.74$)
(N/A) $1$. The initial $pH$ of $0.1 \ M$ $CH_3COOH$ is calculated using the formula: $pH = \frac{1}{2}(pK_a - \log C)$.
$2$. Substituting the values: $pH = \frac{1}{2}(4.74 - \log 0.1) = \frac{1}{2}(4.74 - (-1)) = \frac{1}{2}(5.74) = 2.87$.
$3$. After adding $0.1 \ M$ $CH_3COONa$,the solution becomes a buffer. The $pH$ is calculated using the Henderson-Hasselbalch equation: $pH = pK_a + \log(\frac{[Salt]}{[Acid]})$.
$4$. Substituting the values: $pH = 4.74 + \log(\frac{0.1}{0.1}) = 4.74 + \log(1) = 4.74 + 0 = 4.74$.
$5$. The change in $pH$ is: $\Delta pH = 4.74 - 2.87 = 1.87$.
$2$. Substituting the values: $pH = \frac{1}{2}(4.74 - \log 0.1) = \frac{1}{2}(4.74 - (-1)) = \frac{1}{2}(5.74) = 2.87$.
$3$. After adding $0.1 \ M$ $CH_3COONa$,the solution becomes a buffer. The $pH$ is calculated using the Henderson-Hasselbalch equation: $pH = pK_a + \log(\frac{[Salt]}{[Acid]})$.
$4$. Substituting the values: $pH = 4.74 + \log(\frac{0.1}{0.1}) = 4.74 + \log(1) = 4.74 + 0 = 4.74$.
$5$. The change in $pH$ is: $\Delta pH = 4.74 - 2.87 = 1.87$.
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