Henderson–Hasselbalch equation: pH = pKa + log(c(NaF)/c(HF)).
pH = -log(6,8·10⁻⁴) + log (0,2 M/0,6 M).
pH = 3,17 - 0,48.
pH = 2,69.
Answer:
pH of buffer is 2.69
Explanation:
According to Henderson-Hasselbalch equation for a buffer consists of an weak acid (HF) and it's conjugate base ([tex]F^{-}[/tex])-
[tex]pH=pK_{a}(HF)+log(\frac{C_{F^{-}}}{C_{HF}})[/tex]
Where C stands for concentration
[tex]pK_{a}=-log(K_{a})[/tex]
[tex]C_{F^{-}}[/tex] in 0.2 m NaF is 0.2 m
[tex]C_{HF}=0.6m[/tex]
So, [tex]pH=-log(6.8\times 10^{-4})+log(\frac{0.2}{0.6})=2.69[/tex]