Question

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 1

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
`pK_a`.

The expression used for the calculation of
`pK_a` 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.