19 mLs of neutralizer to neutralize 100 mLs of acid lake water, if we needed to neutralize 50 liters of drinking water, how many grams ?

Only the information on the titration of lake water was provided, but there is no information on that of the drinking acid. Hence, the answer is "unknown".

19ml of neutralizer neutralized 100ml of water. The volume of neutralizer neede to produce 50L of water would be (50,000 x 19)/100 = 9500mL of neutralizer.

(50 L lake water) x (19 mL neutralizer / 0.100 L lake water) = 9500 mL neutralizer

The number of grams depends on the density, which was not given. However, if we take the density given in your previous question:

(9500 mL) x (3 g / 100 mL) = 285 g