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 = $K_a=6.8\times 10^{-4}$

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:

$pK_a=-\log (6.8\times 10^{-4})$

$pK_a=4-\log (6.8)$

$pK_a=3.17$

Now we have to calculate the concentration of NaF.

Using Henderson Hesselbach equation :

$pH=pK_a+\log \frac{[Salt]}{[Acid]}$

$pH=pK_a+\log \frac{[NaF]}{[HF]}$

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

$2.60=3.17+\log (\frac{[NaF]}{0.0310})$

$[NaF]=0.00834M$

Now we have to calculate the moles of NaF.

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

Now we have to calculate the mass of NaF.

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

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