Questions: What is the equilibrium constant for the acid-base reaction between ethanol and acetic acid? The pKs of acetic acid is 4.756, and the pKa of the ethyloxonium ion is -2.4. Be sure your answer has the correct number of significant figures.

What is the equilibrium constant for the acid-base reaction between ethanol and acetic acid? The pKs of acetic acid is 4.756, and the pKa of the ethyloxonium ion is -2.4. Be sure your answer has the correct number of significant figures.
Transcript text: What is the equilibrium constant for the acid-base reaction between ethanol and acetic acid? The $\mathrm{p} K_{s}$ of acetic acid is 4.756, and the $\mathrm{p} K_{a}$ of the ethyloxonium ion is -2.4. Be sure your answer has the correct number of significant figures.
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Solution

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Solution Steps

Step 1: Understanding the Reaction

The reaction between ethanol and acetic acid can be represented as: \[ \text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{C}_2\text{H}_5\text{OH}_2^+ \]

Step 2: Relating pKa and pKs

The equilibrium constant \( K \) for the reaction can be found using the relationship between the pKa values of the acids involved. The equilibrium constant for the reaction is given by: \[ K = 10^{\text{p}K_a(\text{ethyloxonium ion}) - \text{p}K_a(\text{acetic acid})} \]

Step 3: Substituting the Given Values

Given: \[ \text{p}K_a(\text{ethyloxonium ion}) = -2.4 \] \[ \text{p}K_a(\text{acetic acid}) = 4.756 \]

Substitute these values into the equation: \[ K = 10^{-2.4 - 4.756} \]

Step 4: Calculating the Equilibrium Constant

Perform the calculation: \[ K = 10^{-7.156} \]

Step 5: Rounding to Significant Figures

The result should be rounded to four significant figures: \[ K = 7.000 \times 10^{-8} \]

Final Answer

\[ \boxed{K = 7.000 \times 10^{-8}} \]

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