Distilled water has an alkalinity of 2.5 (to pH 4.3). If the water you have has an alkalinity of 10 then it is not very pure having a bicarbonate content of 9 mg/L (assuming there is no other source of alkalinity beyond bicarbonate and hydroxyl ions).
If the 2.5 < alkalinity < 10 you won't need to acidify sparge water as it's buffering capacity is unappreciable compared to the buffering capacity of the mash/wort and pH will not rise above 6 until runoff gravity is very low (beyond the point where it is practical to keep collecting it).
To give you a rough idea as to how much acid is needed to change the pH should you insist on doing so anyway remember that the alkalinity number is the number of ml of normal (1 N) hydrochloric (or any other strong) acid required to move a liter of the water to pH to 4.3 multiplied by 50. Thus for an alkalinity of 10 you would require 10/50 = 0.2 mL of normal acid. Typical hydrochloric acid is approximately 12.4 N so that for 5 gallons of water you would need 19*.2/12.4 = 0.3 mL - not much at all. For DI water you would need 1/4 that much. These additions would bring you to pH 4.3 rather than 6 or 5.8 or something you might consider more suitable for sparge water but again, because of the low buffering capacity, that doesn't really matter.
With phosphoric things are more difficult to compute but as Martin has noted there is a set of formulae at www.wetnewf.org which you can punch into a spreadsheet but then he has already done this for you.
If the 2.5 < alkalinity < 10 you won't need to acidify sparge water as it's buffering capacity is unappreciable compared to the buffering capacity of the mash/wort and pH will not rise above 6 until runoff gravity is very low (beyond the point where it is practical to keep collecting it).
To give you a rough idea as to how much acid is needed to change the pH should you insist on doing so anyway remember that the alkalinity number is the number of ml of normal (1 N) hydrochloric (or any other strong) acid required to move a liter of the water to pH to 4.3 multiplied by 50. Thus for an alkalinity of 10 you would require 10/50 = 0.2 mL of normal acid. Typical hydrochloric acid is approximately 12.4 N so that for 5 gallons of water you would need 19*.2/12.4 = 0.3 mL - not much at all. For DI water you would need 1/4 that much. These additions would bring you to pH 4.3 rather than 6 or 5.8 or something you might consider more suitable for sparge water but again, because of the low buffering capacity, that doesn't really matter.
With phosphoric things are more difficult to compute but as Martin has noted there is a set of formulae at www.wetnewf.org which you can punch into a spreadsheet but then he has already done this for you.