Find the pH when 0.2 , mol of HCl is added to | vedclass

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(D) The added $H^+$ ions from $HCl$ react with $CH_3COO^-$ to form $CH_3COOH$.Initial concentrations: $[CH_3COOH] = 1 \, M$,$[CH_3COO^-] = 1 \, M$.Aft

Vedclass · Accepted Answer

$4.56$

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(D) The added $H^+$ ions from $HCl$ react with $CH_3COO^-$ to form $CH_3COOH$.Initial concentrations: $[CH_3COOH] = 1 \, M$,$[CH_3COO^-] = 1 \, M$.After adding $0.2 \, mol$ of $HCl$:$[CH_3COOH]_{new} = 1 + 0.2 = 1.2 \, M$$[CH_3COO^-]_{new} = 1 - 0.2 = 0.8 \, M$Using the Henderson-Hasselbalch equation:$pH = pK_a + \log \frac{[Salt]}{[Acid]}$$pK_a = -\log(1.8 	imes 10^{-5}) = 5 - \log(1.8) = 5 - 0.255 = 4.745$$pH = 4.745 + \log \frac{0.8}{1.2} = 4.745 + \log(0.667)$$pH = 4.745 - 0.176 = 4.569 \approx 4.56$

Find the $pH$ when $0.2 \, mol$ of $HCl$ is added to $1 \, L$ of a solution containing $1 \, M$ $CH_3COOH$ and $1 \, M$ $CH_3COO^-$. Assume the total volume remains $1 \, L$. Given $K_a$ for $CH_3COOH = 1.8 \times 10^{-5}$.

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