A buffer that is a mixture of acetic acid (K_ | vedclass

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(B) The Henderson-Hasselbalch equation for an acidic buffer is: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$Given: $K_a = 2 	imes 10^{-5}$,so $pK_a = -\l

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$3 : 1$

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(B) The Henderson-Hasselbalch equation for an acidic buffer is: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$Given: $K_a = 2 	imes 10^{-5}$,so $pK_a = -\log(2 	imes 10^{-5}) = 5 - \log 2 = 5 - 0.301 = 4.699 \approx 4.7$.Substituting the values into the equation:$5.18 = 4.7 + \log \frac{[CH_3COO^{-}]}{[CH_3COOH]}$$0.48 = \log \frac{[CH_3COO^{-}]}{[CH_3COOH]}$Since $\log 3 \approx 0.477$,we have $\log 3 \approx 0.48$.Therefore,$\frac{[CH_3COO^{-}]}{[CH_3COOH]} \approx 3 : 1$.

$A$ buffer that is a mixture of acetic acid $(K_a = 2 \times 10^{-5})$ and potassium acetate has $pH = 5.18$. The $\frac{[CH_3COO^{-}]}{[CH_3COOH]}$ ratio in this buffer is approximately:

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