(a) Using the expression Ka=[H+][A−][HA] , explain how to determine which solution has the lower pH , 0.10MHF(aq) or 0.10MHC2H3O2(aq) . Do not perform any numerical calculations.

Ka=[H+][A−]/[HA]

which solution has the lower pH, 0.10MHF(aq) or 0.10MHC2H3O2(aq) :

Higher value of Ka indicates greater ionization thus more protons in solution, higher H+ concentration in solution means lower pH value.

Since HF have higher Ka have, thus HF solution will have lower pH.

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.