But why must we put the OP and the general audience at large through this level of chemistry lesson exercise? And lastly, would you have approached it from this perspective?
Because this is the Brewing Science forum to which people come, presumably, to learn something about brewing science. There are plenty of other fora in which the "I don't care how it works, just tell me what to do" type of questions can be fielded.
Given that we are where we are let's discuss the chemistry correctly:
If we add m moles of NaHCO3 to water two things happen
1) The NaHCO3 dissociates into Na+ ions and HCO3- ions
2)Some of the HCO3- ions give up a proton (H+) and in so doing convert to CO3--
3)Some of the HCO3- ions take up a proton and in so doing convert to H2CO3
4)Most of the H2CO3 becomes CO2 gas and water.
m NaHCO3 + ? H+ <--> m Na+ +m*a2 CO3-- + m*a1 HCO3- + m*a0 H2CO3 <--> m Na+ +m*a2 CO3-- + m*a1 HCO3- + m*a0 CO2 + m*a0 H2O
a0, a1 and a2 are three positive numbers that sum to 1. Thus they represent, respectively, the fractions of the CO moieties that, at the conclusion of the reaction, reside in, respectively, carbonic acid or carbon dioxide, in bicarbonate ion and in carbonate ion.
When we started the charge on the NaHCO3 is 0 made up of m + charges on the sodium ions and m - charges on the bicarbonate. When we are finished there are still m + charges on the sodium but 2*m*a2 on carbonate and m*a1 on bicarbonate. Thus the charge on the carbo moieties has changed from -m to -m*(2*a2 + a1) i.e. by m*(a1 + 2*a2 -1). a0, a1 and a23 depend on the pH at which the mixture of bicarbonate and lactic acid finally arrives. At mash pH a1 is a small number and a1 smaller still thus the charge on bicarbonate has become approximately 1 - a1
less negative. Thus the equation for the sodium bicarbonate is
m NaHCO3 + m*(1 - a1 - 2*a2) H+ <--> m Na+ +m*a2 CO3-- + m*a1 HCO3- + m*a0 CO2 + m*a0 H2O
Clearly the m*(1 - a1 - 2*a2) moles of H+ ions (protons) must come from the lactic acid. If n moles of lactic acid be added to water the following reaction takes place
n HLac <--> n*b1 H+ n*b1 Lac- + n*b0 Hlac.
Here b1 and b0 are again fractions of the totality of Lac. b0 + b1 = 1 and again the b's depend on the final pH of the mix. Note that at mash pH b1 is close to, but not equal, to 1. Thus to neutralize (where "neutralize" means to bring to some desired pH, that being the pH from which the a's and b's are calculated, we require that n*b1 = m*(1 - a1 - 2*a2) meaning that
n = m*(1 - a1 - 2*a2)/b1
moles of lactic acid are required to "neutralize" m moles of bicarbonate. As b1 is close to 1 and as a1 and a2 are small at mash pH it is approximately true that one mole of lactic acid "neutralizes" 1 mole of NaHCO3. Using the molecular weights of NaHCO3 along with the density of HLac solutions of various strengths we can work out the volume of HLac solution of given concentration required to neutralize a gram of NaHCO3.
Note: This may seem a bit complicated at first but we sumplify it by wrtiting spreadsheet functions (or using spreadsheet cells) to calculate Qcarb(pH) and Qlac(pH) where each equals the charge on the anions of the respective acids at the passed value of pH. Assuming we want pHz we know that the charge on the carbo ions changes from -m to m*Qcarb(pHz) i.e. it increases by m*(1 + Qcarb(pHz)) and that the charge on the lactic anion changes from 0 to n*Qlac(pHz) i.e. decreases by -Qlac(pHz). The Q's, representing anion charges, are negative numbers. Thus we immediately calculate
n = -m*((1 + Qcarb(pHz))/Qlac(pHz)
The use of Q's makes the calculations for problems of this nature (of which there are many in brewing) very simple.
Q is computed from pHz using the method given in the Sticky on carbonates.
...provided that the reaction goes to completion, which it eventually should if all of the generated CO2 evolves out of the solution as a gas. There will be of course some portion of CO2 which will not be liberated from solution as a gas, and as such, some degree in which the reaction can potentially be driven in reverse (until such time as nigh on all CO2 is released from the system as a gas).
It should be clear from the foregoing that the reaction will not complete. At mash pH some 8% or so of the added bicarbonate will remain as bicarbonate.