How many moles of HF (Ka=6.8×10−4) must be added to water to form 0.250 L of solution with a pH of 2.90?

The number of moles of HF solution is 8.91 X10⁻⁴ mol

Explanation:

HF -------> H⁺ + F⁻

pH = -Log[H⁺]

α(0.001259) = 6.8 X 10⁻⁴ -  (6.8 X 10⁻⁴)α

(0.001939)α = 0.00068

α = 0.00068/0.001939

α = 0.3507

Recall that, , H⁺ = 0.001259 and Cm = 4n(HF)

Therefore, the number of moles of HF solution that must be added to water to form 0.250 liters of solution with a pH of 2.9 is 8.91 X10⁻⁴ mol

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.