Not a chemist, needs help with a calculation regarding addition of HCl to sodium formate solution As stated I am not a chemist, but I need help on a question I have.

Assume I have a Formic acid 75% and I add between 5-10 kg to 1.000 liter of water. If I wanted to add sodium formate and HCl 37% to water and achieve a similar result  with regards to pH and "pure" undissociated formic acid how much of each product would I need to add?

I have calculated, probaly wrong, that I would need 8 ml of HCl to 1.000 L to lower pH to around 4 which is close to pKa of FA (3,75), but I do not know if this will give me the same amount of undissociated FA as the original scenario.

For my application idea pH is less important than undissociated (antibacterial) FA.

Hope someone takes up the (over my head) calculation challenge
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% $\ce{HCl}$