Question

How many moles of solid NaF would have to be added to 1.0 L of 0.115 M HF solution to achieve a buffer of pH 3.46? Assume there is no volume change. Ka for HF = 7.2 × 10 –4

Answer 1

Answer : The  moles of solid NaF is, 1.09 mole.

Explanation : Given,

pH = 3.46

$K_a=7.2\times 10^{-4}$

Concentration of HF = 0.115 M

Volume of solution = 1.0 L

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:

$pK_a=-\log (7.2\times 10^{-4})$

$pK_a=4-\log (7.2)$

$pK_a=3.14$

Now we have to calculate the moles of HF.

$\text{Moles of }HF=\text{Concentration of }HF\times \text{Volume of solution}$

$\text{Moles of }HF=0.115M\times 1.0L=0.115mol$

Now we have to calculate the moles of NaF.

Using Henderson Hesselbach equation :

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

$pH=pK_a+\log \frac{\text{Moles of NaF}}{\text{Moles of HF}}$

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

$3.46=3.14+\log (\frac{\text{Moles of NaF}}{0.115})$

$\text{Moles of NaF}=1.09mol$

Thus, the moles of solid NaF is, 1.09 mole.