Calculate volume of $0,1 \mathrm{M}$ ethanoic acid $\mathrm{CH} 3 \mathrm{COOH}$ and $0,1 \mathrm{M}$ sodium ethanoate $\mathrm{CH} 3$ COONa needed to prepare $100 \mathrm{ml}$ of buffer with $\mathrm{pH} \mathrm{4,50!} \mathrm{Please} \mathrm{kindly} \mathrm{explain} \mathrm{and}$ show(!!) how do the calculations! Thanks so much!

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∻Solution∻ ‖ 1 ⋮ **Use the Henderson-Hasselbalch equation**: The pH of a buffer solution can be calculated using the Henderson-Hasselbalch equation, which is given by $ ext{pH} = ext{pKa} + \log \left( \frac{[ ext{A}^-]}{[ ext{HA}]} ight)$, where $ ext{pKa}$ is the acid dissociation constant, $[ ext{A}^-]$ is the concentration of the conjugate base, and $[ ext{HA}]$ is the concentration of the acid. ‖ ‖ 2 ⋮ **Calculate the ratio of concentrations**: For ethanoic acid, $ ext{pKa} = 4.75$. Given that the desired pH is 4.50, we can rearrange the Henderson-Hasselbalch equation to solve for the ratio $\frac{[ ext{CH}_3 ext{COO}^-]}{[ ext{CH}_3 ext{COOH}]}$: $ ext{pH} = ext{pKa} + \log \left( \frac{[ ext{CH}_3 ext{COO}^-]}{[ ext{CH}_3 ext{COOH}]} ight)$, which gives $\log \left( \frac{[ ext{CH}_3 ext{COO}^-]}{[ ext{CH}_3 ext{COOH}]} ight) = ext{pH} - ext{pKa} = 4.50 - 4.75 = -0.25$. Therefore, $\frac{[ ext{CH}_3 ext{COO}^-]}{[ ext{CH}_3 ext{COOH}]} = 10^{-0.25}$. ‖ ‖ 3 ⋮ **Determine the volumes of acid and conjugate base**: Since the total volume of the buffer is 100 mL and the concentrations of both the acid and the conjugate base are 0.1 M, we can use the ratio from step 2 to find the volumes. Let $V_{ ext{acid}}$ be the volume of ethanoic acid and $V_{ ext{base}}$ be the volume of sodium ethanoate. We have $V_{ ext{acid}} + V_{ ext{base}} = 100$ mL and $\frac{V_{ ext{base}}}{V_{ ext{acid}}} = 10^{-0.25}$. Solving these equations gives us $V_{ ext{acid}} = \frac{100}{1 + 10^{-0.25}}$ mL and $V_{ ext{base}} = 100 - V_{ ext{acid}}$ mL. ‖ ‖ 4 ⋮ **Calculate the exact volumes**: Plugging the values into the equations from step 3, we get $V_{ ext{acid}} \approx 56$ mL and $V_{ ext{base}} \approx 44$ mL. ‖ ∻Answer∻ ⚹ To prepare 100 mL of a buffer with pH 4.50 using 0.1 M ethanoic acid and 0.1 M sodium ethanoate, you need approximately 56 mL of ethanoic acid and 44 mL of sodium ethanoate. ⚹ ∻Key Concept∻ ⚹ Henderson-Hasselbalch equation ⚹ ∻Explanation∻ ⚹ The Henderson-Hasselbalch equation relates the pH of a buffer solution to the pKa of the acid and the ratio of the concentrations of the conjugate base and the acid. By rearranging this equation, we can calculate the necessary volumes of acid and conjugate base to create a buffer with a specific pH. ⚹◊What is the Henderson-Hasselbalch equation and how can it be used to calculate the volumes of ethanoic acid and sodium ethanoate needed to prepare a buffer solution with a pH of 4.50?⍭ Generate me a similar question◊

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