Perhaps another way to gain some perspective on this is to consider a water alkaline enough that a base malt mash comes in at pH 5.8. Since it is claimed that 5.2 will bring the pH of this mash to 5.2 we note that a drop of 0.6 pH is required of the product. As most mashes have a pound of grain per quart or so we will assume .5 kg/L (just to make the math easier). And we will also assume the malt has a buffering capacity of 25 mEq/Kg-ph which is not an unreasonable assumption. Thus 5.2 has to provide 0.6*0.5*25 = 7.5 mEq of H+ per liter. We know that 5.2 is essentially monobasic sodium phosphate. If we assumed that all the sodium phosphate released its proton it would then require 7.5 mEq/L sodium phosphate to move this mash to 5.2. That is 7.5*136 = 1020 grams of the salt of which 7.5*23 = 172.5 mg is sodium. There is no way I would consider adding over a gram of this product to each liter of my water especially knowing that it would result in 175 mg/L sodium. And especially when there are simpler, more effective, more predictable means at hand i.e. dilution with RO and addition of sauermalz.