Step 1: Calculate the $\text{p}K_a$ of formic acid. The acid dissociation constant ($K_a$) for formic acid (HCOOH) is given as $1.77 \times 10^{-4}$. $$ \text{p}K_a = -\log_{10}(K_a) $$ $$ \text{p}K_a = -\log_{10}(1.77 \times 10^{-4}) $$ $$ \text{p}K_a \approx 3.752 $$ Step 2: Use the Henderson-Hasselbalch equation to find the required concentration of the conjugate base ($[\text{HCOO}^-]$). The Henderson-Hasselbalch equation is: $$ \text{pH} = \text{p}K_a + \log_{10}\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) $$ We are given: $\text{pH} = 3$ $[\text{HA}] = [\text{HCOOH}] = 0.50 \text{ M}$ $\text{p}K_a = 3.752$ Substitute these values into the equation: $$ 3 = 3.752 + \log_{10}\left(\frac{[\text{HCOO}^-]}{0.50}\right) $$ Subtract $3.752$ from both sides: $$ 3 - 3.752 = \log_{10}\left(\frac{[\text{HCOO}^-]}{0.50}\right) $$ $$ -0.752 = \log_{10}\left(\frac{[\text{HCOO}^-]}{0.50}\right) $$ To remove the logarithm, take $10$ to the power of both sides: $$ 10^{-0.752} = \frac{[\text{HCOO}^-]}{0.50} $$ $$ 0.177 = \frac{[\text{HCOO}^-]}{0.50} $$ Now, solve for $[\text{HCOO}^-]$: $$ [\text{HCOO}^-] = 0.177 \times 0.50 \text{ M} $$ $$ [\text{HCOO}^-] = 0.0885 \text{ M} $$ Step 3: Calculate the moles of sodium formate (HCOONa) needed. The volume of the buffer solution is $500 \text{ mL}$, which is $0.500 \text{ L}$. Moles of HCOONa = Concentration $\times$ Volume $$ \text{Moles of HCOONa} = 0.0885 \text{ mol/L} \times 0.500 \text{ L} $$ $$ \text{Moles of HCOONa} = 0.04425 \text{ mol} $$ Step 4: Calculate the mass of sodium formate (HCOONa) needed. The molar mass of sodium formate (HCOONa) is $68.0069 \text{ g/mol}$. Mass = Moles $\times$ Molar mass $$ \text{Mass of HCOONa} = 0.04425 \text{ mol} \times 68.0069 \text{ g/mol} $$ $$ \text{Mass of HCOONa} \approx 3.009 \text{ g} $$ The mass of sodium formate needed is approximately $3.01 \text{ g}$. The final answer is $\boxed{\text{3.01 g}}$. What's next?