Calculate the mass of sodium acetate that must be added to 500.0 \(\mathrm{mL}\) of 0.200\(M\) acetic acid to form a \(\mathrm{pH}=5.00\) buffer solution.

First, we need to find the pKa of acetic acid. Acetic acid is a weak acid with a known dissociation constant, \(K_a = 1.8 \times 10^{-5}\). We can find its pKa by taking the negative logarithm of its \(K_a\) value: \( pK_a = -\log{K_a} = -\log{(1.8 \times 10^{-5})} \) Calculating this yields: \( pK_a \approx 4.74 \)

Now that we have the pKa, we can use the Henderson-Hasselbalch equation to determine the concentration of the conjugate base, sodium acetate: \(pH = pK_a + \log{\frac{[A^-]}{[HA]}} \) We are given the desired pH (5.00) and the initial concentration of acetic acid (\(0.200 M\)). The concentration of the conjugate base (\([A^-]\)) is unknown. Plugging the given values into the equation: \(5.00 = 4.74 + \log{\frac{[A^-]}{0.200}} \) Now we will solve for \([A^-]\).

To isolate \([A^-]\), we first need to subtract the pKa from both sides of the equation: \( 5.00 - 4.74 = \log{\frac{[A^-]}{0.200}} \) \( 0.26 = \log{\frac{[A^-]}{0.200}} \) Next, we will exponentiate both sides of the equation with base 10 to remove the logarithm: \( 10^{0.26} = \frac{[A^-]}{0.200} \) Now, we will multiply both sides of the equation by 0.200 to solve for \([A^-]\): \( [A^-] = 10^{0.26} \times 0.200 \) Calculating this yields: \( [A^-] \approx 0.292 M \)

We have the required concentration of sodium acetate (0.292 M), and we know the volume of the solution (500.0 mL). Now we can calculate the moles of sodium acetate needed, then convert it to mass using its molar mass: First, convert the volume to liters: \( Volume = 500.0 \, mL \times \frac{1 \, L}{1000 \, mL} = 0.500 \, L \) Next, calculate the moles of sodium acetate needed for the buffer solution: \( Moles \, of \, sodium \, acetate = [A^-] \times Volume = 0.292 \, M \times 0.500 \, L \) Calculating this yields: \( Moles \, of \, sodium \, acetate \approx 0.146 \, mol \) Finally, convert moles of sodium acetate to mass using its molar mass (\(Molar \, mass \, of \, sodium \, acetate = 82.03 \, g/mol\)): \[Mass \, of \, sodium \, acetate = Moles \times Molar \, mass = 0.146 \, mol \times 82.03 \, g/mol \] Calculating this yields: \( Mass \, of \, sodium \, acetate \approx 11.98 \, g \) Thus, approximately 11.98 grams of sodium acetate must be added to 500.0 mL of 0.200 M acetic acid to form a pH=5.00 buffer solution.