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The Henderson-Hasselbalch equation is given by: $$ \mathrm{pH} = \mathrm{p}K_{a} + \log\left(\frac{\left[\mathrm{A^-}\right]}{\left[\mathrm{HA}\right]}\right) $$ In this equation, pH is the desired pH of the buffer, pKa is the negative logarithm of the acid dissociation constant, and the fraction inside the logarithm represents the ratio of the concentrations of the conjugate base (A-) to the acid (HA).

The acid dissociation constant (Ka) for bromoacetic acid can be found in a table or online. For bromoacetic acid, the Ka value is \(1.3 \times 10^{-2}\). To find the pKa value, take the negative logarithm of Ka: $$ \mathrm{p}K_{a} = -\log(K_{a}) = -\log(1.3 \times 10^{-2}) = 1.89 $$

Now that we have the pKa value, we can plug it into the Henderson-Hasselbalch equation along with the desired pH and solve for the ratio of the concentrations of the conjugate base and the acid: $$ 3.00 = 1.89 + \log\left(\frac{\left[\mathrm{A^-}\right]}{\left[\mathrm{HA}\right]}\right) $$ Solving for the ratio: $$ \frac{\left[\mathrm{A^-}\right]}{\left[\mathrm{HA}\right]} = 10^{(3.00-1.89)} = 10^{1.11} = 12.92 $$

Now that we have the ratio of the concentrations, we can express the total concentration of the two components (0.200 M) using the ratio: $$ \left[\mathrm{A^-}\right] + \left[\mathrm{HA}\right] = 0.200\,\mathrm{M} $$ With the ratio we found in Step 3: $$ \left[\mathrm{A^-}\right] = 12.92\,\left[\mathrm{HA}\right] $$ We can now solve for the individual concentrations: $$ (1 + 12.92)\left[\mathrm{HA}\right] = 0.200\,\mathrm{M} $$ $$ \left[\mathrm{HA}\right] = \frac{0.200\,\mathrm{M}}{13.92} = 0.0144\,\mathrm{M} $$ $$ \left[\mathrm{A^-}\right] = 12.92 \times 0.0144\,\mathrm{M} = 0.186\,\mathrm{M} $$

Now that we have the individual concentrations, we can calculate the masses of the two components needed to prepare 1.00 L of the buffer. We will first find the moles of each component and then convert it into masses using their respective molar masses. Moles of bromoacetic acid (HA) = concentration × volume $$ \mathrm{moles}\,\mathrm{HA} = 0.0144\,\mathrm{M} \times 1\,\mathrm{L} = 0.0144\,\mathrm{mol} $$ Moles of sodium bromoacetate (A-) = concentration × volume $$ \mathrm{moles}\,\mathrm{A^-} = 0.186\,\mathrm{M} \times 1\,\mathrm{L} = 0.186\,\mathrm{mol} $$ Using the molar masses of bromoacetic acid (138.94 g/mol) and sodium bromoacetate (188.95 g/mol), we can find the masses: Mass of bromoacetic acid, HA (g) = moles × molar mass $$ \mathrm{mass}\,\mathrm{HA} = 0.0144\,\mathrm{mol} \times 138.94\,\mathrm{g/mol} = 2.00\,\mathrm{g} $$ Mass of sodium bromoacetate, A- (g) = moles × molar mass $$ \mathrm{mass}\,\mathrm{A^-} = 0.186\,\mathrm{mol} \times 188.95\,\mathrm{g/mol} = 35.14\,\mathrm{g} $$ Therefore, to prepare 1.00 L of pH = 3.00 buffer with a total concentration of 0.200 M, 2.00 g of bromoacetic acid and 35.14 g of sodium bromoacetate are needed.