Be sure to answer all parts. tris [tris(hydroxymethyl)aminomethane] is a common buffer for studying biological systems. (k=5.01 x 10 and pk, = 8.3) (a) calculate the ph of the tris buffer after mixing 10.0 ml of 0.20 m hcl solution with 25.0 ml of 0.10 m tris. (b) this buffer was used to study an enzyme-catalyzed reaction. as a result of the reaction, 0.00019 mol of h was produced. what is the ph of the buffer at the end of the reaction? present? (c) what would be the final ph if no buffer were

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.