NaOH being added to Formic Acid (Please catch my error!)
It looks like you misused the acid dissociation constant, which is apparently the $K_a=1.80×10^{-4}$ figure you have.

Properly, you should do the following:

1)  Let $x$ be the hydrogen ion concentration $[\ce{H^+}]$ from the formic acid that actually disdociated.

2)  Since you ended up with $0.469 \text{M}$ total acid the un-dissociated acid $[\ce{HCOOH}]=0.469-x$.

3)  The formate ion comes from both the salt ($0.00521 \text{M}$) and the dissociated acid ($x$), so $[\ce{HCOO^-}]=0.00521+x$.

4)  Plug these into the equilibrium relation

$\frac{[\ce{H^+}][\ce{HCOO^-}]}{[\ce{HCOOH}]}=K_a$

and solve for $x$.  The equation reduces down to a quadratic equation so you should have an easy formula for $x=[\ce{H^+}]$.

5)  Get the pH from the result of Step 4.

Good luck!