Calculate the pK_a of formic acid.

The acid dissociation constant ($K_a$) for formic acid (HCOOH) is given as $1.77 \times 10^{-4}$.

$$\text{p}K_a = -\log_{10}(K_a)$$ $$\text{p}K_a = -\log_{10}(1.77 \times 10^{-4})$$ $$\text{p}K_a \approx 3.752$$

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 $[\text{HCOO}^-]$:

$$[\text{HCOO}^-] = 0.177 \times 0.50 \text{ M}$$ $$[\text{HCOO}^-] = 0.0885 \text{ M}$$

The volume of the buffer solution is $500 \text{ mL}$, which is $0.500 \text{ L}$.
Moles of HCOONa = Concentration $\times$ Volume

$$\text{Moles of HCOONa} = 0.0885 \text{ mol/L} \times 0.500 \text{ L}$$ $$\text{Moles of HCOONa} = 0.04425 \text{ mol}$$

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 $3.01 \text{ g}$.