How many grams of sodium fluoride should be added to 300. ml of 0.0310 m of hydrofluoric acid to produce a buffer solution with a ph of 2.60?

Answer : The mass of sodium fluoride added should be 0.105 grams.

Explanation : Given,

The dissociation constant for HF =

Concentration of HF (weak acid)= 0.0310 M

First we have to calculate the value of .

The expression used for the calculation of is,

Now put the value of in this expression, we get:

Now we have to calculate the concentration of NaF.

Using Henderson Hesselbach equation :

Now put all the given values in this expression, we get:

Now we have to calculate the moles of NaF.

Now we have to calculate the mass of NaF.

Therefore, the mass of sodium fluoride added should be 0.105 grams.

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.

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