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?

Question

asked May 1, 2020 8.0k views

1 Answer

Tuxdna · Answer 1 · 2020-05-07T11:10:58+0000

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

Explanation : Given,

The dissociation constant for HF =
K_a=6.8* 10^(-4)

Concentration of HF (weak acid)= 0.0310 M

First we have to calculate the value of
.

The expression used for the calculation of
is,

$pK_a=-\log (K_a)$

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

$pK_a=-\log (6.8* 10^(-4))$

$pK_a=4-\log (6.8)$

pK_a=3.17

Now we have to calculate the concentration of NaF.

Using Henderson Hesselbach equation :

$pH=pK_a+\log ([Salt])/([Acid])$

$pH=pK_a+\log ([NaF])/([HF])$

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

$2.60=3.17+\log (([NaF])/(0.0310))$

[NaF]=0.00834M

Now we have to calculate the moles of NaF.

$\text{Moles of NaF}=\text{Concentration of NaF}* \text{Volume of solution}=0.00834M* 0.300L=0.0025mole$

Now we have to calculate the mass of NaF.

$\text{Mass of }NaF=\text{Moles of }NaF* \text{Molar mass of }NaF=0.0025mole* 42g/mole=0.105g$

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