A 0.200 M NaOH solution was used to titrate a 18.25 mL HF solution. The endpoint was reached after 31.20 mL of titrant
were added. Find the molar concentration of the original HF
solution.

The molar concentration of HF is 0.34 M

Explanation:

Given data:

Molarity of NaOH =M₁ =  0.200 M

Volume of HF =V₂=  18.25 mL

Volume of NaOH added = V₁ = 31.20 mL

Molarity of HF solution = M₂ = ?

Solution:

M₁V₁  =  M₂V₂

by putting values,

0.200 M × 31.20 mL = M₂ × 18.25 mL

M₂ = 0.200 M × 31.20 mL /  18.25 mL

M₂ = 6.24 M.mL / 18.25 mL

M₂ = 0.34 M

Thus, the molar concentration of HF is 0.34 M.

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.

> b takes place in the gas phase in an isothermal tubular reactor. the feed is one mole of a per one mole of c, an inert. the entering temperature and pressure are 427°c and 10 atm, respectively. the gas constant r = 0.08206 atm · l/mol · k.