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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?

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

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In a mixture of two acids, nothing happens : no reaction between the acids. No salt !

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.

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