pH calculation of HCOOH +plus HCl solution

Question - What volume of $$0.1 \ce{M}$$ $$\ce{HCl}$$ solution should be added to a $$\ce{500 mL}$$ of $$\ce{0.5 M }$$ $$\ce{HCOOH }$$ solution in order to prepare an acid solution of $$\text{pH}$$ =$$1.5.$$ $$K_a$$ of formic acid is $$2×10^{-4}$$ at $$\ce{298 K?}$$

I am not sure how to proceed with this question. Iam also confused how equilibrium reaction will

I guessed it would be $$\ce{HCOOH}+\ce{HCl \rightarrow salt}$$ Let $$V$$ be volume of HCl. At $$t=\text{equilibrium,}$$ $$[\ce{salt}] = \dfrac{0.1V}{0.25-0.1V}$$

Now, $$\ce{pH}= \ce{pK_a} + \log\dfrac{0.1V}{(0.25-0.1V)}\Rightarrow \boxed{V=2.5L}$$ But answer is $$V=0.208 L$$

Am I completely wrong somewhere? Can someone give tell me how to proceed?

$$\textbf{NOTE:I am not asking full solution.}$$ $$\textbf{Just telling how to proceed is fine.}$$

At $$\ce{pH = 1.5}$$, the finall concentration of $$\ce{H+}$$ is $$\ce{10^{-1.5} = 0.0316}$$ M. This concentration is the sum of the $$\ce{H+}$$ ions produced by $$\ce{HCl}$$ and those from $$\ce{HCOOH}$$. The amount of $$\ce{HCOOH}$$ is $$0.5$$ mol/L·$$0.5$$ L$$= 0.25$$ mol. If the concentration of formate ion is $$x$$, $$x$$ is also the concentration of dissociated $$\ce{HCOOH}$$ molecules. Now the dissociation constant of formic acid is $$\ce{K_a = \frac{0.0316·x}{0.25 - x} = 2·10^{-4}}$$, From there, $$x$$ = $$1.6·10^{-3}$$.
The concentration of $$\ce{H+}$$ ions coming form $$\ce{HCl}$$ is $$0.0316 - 0.0016 = 0.0300$$ M. The total volume of the final solution is : $$0.5 + \ce{V}$$, it V is the volume of $$\ce{HCl}$$ to add. The amount in moles of $$\ce{HCl}$$ to be added is $$0.0300·(\ce{0.5 + V)}$$. The volume V where to find this amount of $$\ce{HCl}$$ is : $$\ce{V = \frac{n}{c}} = \frac{0.03·(0.5 + V)}{0.1}$$ . From there, one gets : $$\ce{V = \frac{0.15}{0.7} = 0.214 L}$$
It is "nearly" your expected value !! If you remake the calculations, with more decimals, you should obtain $$0.208$$ L.