1.0 mol of HCl and 1.0 mol of CH_3COONa are | vedclass

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(A) The reaction between $HCl$ and $CH_3COONa$ is: $CH_3COONa + HCl \rightarrow CH_3COOH + NaCl$. Since $1.0 \ mol$ of each reactant is used,they reac

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$4 	imes 10^{-3}$

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(A) The reaction between $HCl$ and $CH_3COONa$ is: $CH_3COONa + HCl ightarrow CH_3COOH + NaCl$. Since $1.0 \ mol$ of each reactant is used,they react completely to form $1.0 \ mol$ of $CH_3COOH$ in $1 \ L$ of solution. The concentration of $CH_3COOH$ is $C = 1.0 \ M$. $CH_3COOH$ is a weak acid that dissociates as: $CH_3COOH ightleftharpoons CH_3COO^- + H^+$. The dissociation constant is $K_a = \frac{[CH_3COO^-][H^+]}{[CH_3COOH]} = \frac{x^2}{C-x} \approx \frac{x^2}{C}$ (where $x = [H^+]$). $x = \sqrt{K_a 	imes C} = \sqrt{1.6 	imes 10^{-5} 	imes 1.0} = \sqrt{16 	imes 10^{-6}} = 4 	imes 10^{-3} \ M$.

$1.0 \ mol$ of $HCl$ and $1.0 \ mol$ of $CH_3COONa$ are dissolved in water to make a $1 \ L$ solution. What is the concentration of $H^+$ in the solution? ($K_a$ for $CH_3COOH = 1.6 \times 10^{-5}$)

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