You are asked to prepare a \(\mathrm{pH}=3.00\) buffer starting from $2.00 \mathrm{~L}\( of \)0.025 \mathrm{M}$ solution of benzoic acid \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\right)\) and any amount you need of sodium benzoate $\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COONa}\right) .(\mathbf{a})\( What is the \)\mathrm{pH}$ of the benzoic acid solution prior to adding sodium benzoate? (b) How many grams of sodium benzoate should be added to prepare the buffer? Neglect the small volume change that occurs when the sodium benzoate is added.

The acid dissociation constant, Ka, is given for benzoic acid: Ka = \(6.3 \times 10^{-5}\). Given the equation: pH = -log[H+] and Ka = \(\frac{[H^+] [A^-]}{[HA]}\), We can write the equation for Ka as: \(6.3 \times 10^{-5} = \frac{[H^+] [A^-]}{[HA]}\), where [HA] is the concentration of benzoic acid, which is 0.025 M, [A^-] is the concentration of the benzoate ion, and [H+] is the hydrogen ion concentration. Since the auto-ionization of water can be neglected initially, we can assume [A^-] = [H+]. Plugging these values into the equation: \(6.3 \times 10^{-5} = \frac{[H^+]^2}{0.025}\), Solve for [H+]: [H+] = \(1.58 \times 10^{-3} \mathrm{M}\). Now calculate the pH: pH = -log(\(1.58 \times 10^{-3}\)) ≈ 2.80.

The Henderson-Hasselbalch equation is given by: pH = pKa + log \(\frac{[A^-]}{[HA]}\), where pKa = -log(Ka). The desired pH of the buffer is 3.00 and we found the pH of benzoic acid solution as 2.80. Therefore: 3.00 = 4.20 + log \(\frac{[A^-]}{0.025}\), Solving for [A^-] (sodium benzoate concentration): [A^-] = \(0.0618 \mathrm{M}\).

To find the required moles of sodium benzoate to prepare the buffer, we can use the equation: Moles of sodium benzoate = [A^-] * volume In this case, the volume is 2.00 L: Moles of sodium benzoate = \(0.0618 \times 2.00 = 0.1236 \mathrm{mol}\). Now, we can find the mass of sodium benzoate to be added by using its molar mass (M = \(144.10 \mathrm{g/mol}\)): Mass of sodium benzoate = moles * molar mass Mass of sodium benzoate = \(0.1236 \times 144.10 = 17.8 \mathrm{g}\). So, 17.8 grams of sodium benzoate need to be added to the benzoic acid solution to prepare a pH 3.00 buffer.