Not a chemist, needs help with a calculation regarding addition of HCl to sodium formate solution
At $\mathrm{pH}=\mathrm{p}K_\mathrm{a}=3.75$ , there will be 50% of non-dissociated formic acid and 50% of formiate. 

You need 50 % of stoichiometric amount of hydrochloric acid to convert 50% of the formiate to the acid.

If we consider the percentage as (w/w) %,  then $\pu{1 kg}$ of 75% formic acid ($M=\pu{46 
g/mol}$) contains $\pu{16.3 mol}$ of formic acid.

For usage of $\pu{5 kg}$ to $\pu{1000 L}$, $\mathrm{pH}=\frac 12 \left(\mathrm{p}K_\mathrm{a}-\log(c)\right)\\
=\frac12\cdot \left(3.75-\log{\frac{5\cdot 16.3}{1000}}\right)\\
=\frac12\cdot \left(3.75-\log{0.0815}\right)=2.42$

For $\pu{10 kg}$, it is 2.27.

$\pu{1 kg}$ of the above formic acid  is equivalent to $\pu{ 16.3 mol}$ of sodium formiate, i.e. $\pu{ 1108 g}$

For 50% conversion, you need $\pu{ 8.15 mol}$ of $\ce{HCl}$, i.e. $\pu{ 8.15\cdot 36.5 = 297 g}$ $\ce{HCl}$, what is equivalent $\pu{ 804 g}$ of 37% $\ce{HCl}$