Respuesta :
Answer : The moles of solid NaF is, 1.09 mole.
Explanation : Given,
pH = 3.46
[tex]K_a=7.2\times 10^{-4}[/tex]
Concentration of HF = 0.115 M
Volume of solution = 1.0 L
First we have to calculate the value of [tex]pK_a[/tex].
The expression used for the calculation of [tex]pK_a[/tex] is,
[tex]pK_a=-\log (K_a)[/tex]
Now put the value of [tex]K_a[/tex] in this expression, we get:
[tex]pK_a=-\log (7.2\times 10^{-4})[/tex]
[tex]pK_a=4-\log (7.2)[/tex]
[tex]pK_a=3.14[/tex]
Now we have to calculate the moles of HF.
[tex]\text{Moles of }HF=\text{Concentration of }HF\times \text{Volume of solution}[/tex]
[tex]\text{Moles of }HF=0.115M\times 1.0L=0.115mol[/tex]
Now we have to calculate the moles of NaF.
Using Henderson Hesselbach equation :
[tex]pH=pK_a+\log \frac{[Salt]}{[Acid]}[/tex]
[tex]pH=pK_a+\log \frac{\text{Moles of NaF}}{\text{Moles of HF}}[/tex]
Now put all the given values in this expression, we get:
[tex]3.46=3.14+\log (\frac{\text{Moles of NaF}}{0.115})[/tex]
[tex]\text{Moles of NaF}=1.09mol[/tex]
Thus, the moles of solid NaF is, 1.09 mole.
