Quantity of base needed to neutralise litre of 36% HCl?

Sorry for asking a basic chemistry question. I last did this kind of chemistry at school about 30 years ago and I don't have confidence that I'd get it right from first principles :)


Short version: Hydrochloric acid is used in some kinds of works to remove cement smears from masonry and paving slabs. But on a level surface, if left there, it's also able to degrade the cement that's *supposed* to be there (cement is highly alkali). Usually you can wash it away with water, but in my situation that's not workable. Instead, I plan to buy some Na2CO3, NaHCO3 or NaOH, all of which are ordinary commodity items at any local shop, dissolve in water, and carefully pour to neutralise any unreacted acid as soon as the surface is clean of cement smears. Excess alkali isn't a problem.


I've used concentrated HCl for many years, and the fumes/reaction heat/safety kit won't be an issue, so I'm comfortable with the safety aspects of this approach - comments not needed on safety.


My question is about making sure I have enough base ready to use, to neutralise the amount of acid I end up using. The acid is already on-hand, it's fuming 36% HCl. I'll dilute it for use, of course, but I can work out how many litres equivalent of 36% ends up being used. The bases (whichever I use) will be dry powder/crystal form. Obviously I pour until no reaction is visible, but if I don't buy enough, I'd have a problem. (Buying a bit more than I need is fine)


So I need a reminder how to work out the mass of each of the above 3 dry/crystalline bases that would be required to neutralise 1 litre of 36% HCl, so I can get an idea what kind of quantity of base to buy, before I start the task.


Given that helping hand with the basic chemical calculations, I can work out the rest easily.




The molarity of 36% HCl is approx 11.7, i.e. there are 11.7 moles per litre. To neutralise the whole litre you need 11.7 moles of NaOH, 11.7 moles of NaHCO3 or 5.85 moles of Na2CO3. Multiply the number of moles by the molecular weight - 40 in the case of NaOH, 106 for Na2CO3, 84 for NaHCO3 - to give you the weight required. 


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