A biochemist needs \(750 \mathrm{~mL}\) of an acetic acid-sodium acetate buffer with \(\mathrm{pH}\) 4.50. Solid sodium acetate \(\left(\mathrm{CH}_{3} \mathrm{COONa}\right)\) and glacial acetic acid \(\left(\mathrm{CH}_{3} \mathrm{COOH}\right)\) are available. Glacial acetic acid is \(99 \% \mathrm{CH}_{3} \mathrm{COOH}\) by mass and has a density of \(1.05 \mathrm{~g} / \mathrm{mL}\). If the buffer is to be \(0.15 \mathrm{M}\) in \(\mathrm{CH}_{3} \mathrm{COOH}\), how many grams of \(\mathrm{CH}_{3} \mathrm{COONa}\) and how many milliliters of glacial acetic acid must be used?

The Henderson-Hasselbalch equation is given as: \[ \mathrm{pH} = \mathrm{p}K_a + \log \frac{[\mathrm{A}^-]}{[\mathrm{HA}]} \] where \([\mathrm{A}^-]\) represents the concentration of the conjugate base (sodium acetate), \([\mathrm{HA}]\) represents the concentration of the weak acid (acetic acid) and 𝑝𝐾𝑎 is the acid dissociation constant of the weak acid. For acetic acid, the 𝑝𝐾𝑎 value is 4.74. Given the pH of the buffer (which is 4.50) and the concentration of acetic acid (0.15 M), we can calculate the concentration of sodium acetate needed as follows: \[ 4.50 = 4.74 + \log \frac{[\mathrm{CH}_3\mathrm{COONa}]}{0.15} \]

Rearrange the equation to solve for the concentration of sodium acetate: \[ \log \frac{[\mathrm{CH}_3\mathrm{COONa}]}{0.15} = 4.50 - 4.74 \] \[ \log [\mathrm{CH}_3\mathrm{COONa}] - \log(0.15) = -0.24 \] \[ [\mathrm{CH}_3\mathrm{COONa}] = 10^{-0.24} \times 0.15 \] Calculating the concentration of sodium acetate, we get: \[ [\mathrm{CH}_3\mathrm{COONa}] = 0.134 \mathrm{M} \]

We know the concentration and the volume of the buffer solution, so we can find the moles and, subsequently, the mass of sodium acetate needed: \[ \mathrm{moles~of~sodium~acetate} = [\mathrm{CH}_3\mathrm{COONa}] \times \mathrm{volume~of~buffer} \] \[ \mathrm{moles~of~sodium~acetate} = 0.134 \mathrm{M} \times 0.750 \mathrm{L} = 0.1005 \mathrm{~mol} \] Next, we will find the mass using the molar mass of sodium acetate (82.03 g/mol): \[ \mathrm{mass~of~sodium~acetate} = \mathrm{moles~of~sodium~acetate} \times \mathrm{molar~mass~of~sodium~acetate} \] \[ \mathrm{mass~of~sodium~acetate} = 0.1005 \mathrm{~mol} \times 82.03 \frac{\mathrm{g}}{\mathrm{mol}} = 8.24 \mathrm{~g} \]

We know the buffer is to be 0.15 M in acetic acid. Convert concentration to moles using the volume of the buffer: \[ \mathrm{moles~of~acetic~acid} = [\mathrm{CH}_3\mathrm{COOH}] \times \mathrm{volume~of~buffer} \] \[ \mathrm{moles~of~acetic~acid} = 0.15 \mathrm{M} \times 0.750 \mathrm{L} = 0.1125 \mathrm{~mol} \] Now, we will find the mass of acetic acid using the molar mass of acetic acid (60.05 g/mol): \[ \mathrm{mass~of~acetic~acid} = \mathrm{moles~of~acetic~acid} \times \mathrm{molar~mass~of~acetic~acid} \] \[ \mathrm{mass~of~acetic~acid} = 0.1125 \mathrm{~mol} \times 60.05 \frac{\mathrm{g}}{\mathrm{mol}} = 6.756 \mathrm{~g} \] Now, let's calculate the volume considering that glacial acetic acid has only 99% acetic acid by mass and density \(1.05 \frac{\mathrm{g}}{\mathrm{mL}}\): \[ \mathrm{mass~of~glacial~acetic~acid} = \frac{\mathrm{mass~of~acetic~acid}}{0.99} = \frac{6.756 \mathrm{~g}}{0.99} = 6.82 \mathrm{~g} \] \[ \mathrm{volume~of~glacial~acetic~acid} = \frac{\mathrm{mass~of~glacial~acetic~acid}}{\mathrm{density}}=\frac{6.82 \mathrm{~g}}{1.05 \frac{\mathrm{g}}{\mathrm{mL}}} = 6.49 \mathrm{~mL} \] In conclusion, to prepare 750 mL of a pH 4.50 buffer with a 0.15 M acetic acid, we must use 8.24 grams of sodium acetate and 6.49 mL of glacial acetic acid.