Answer:
The correct option is d
Step-by-step explanation:
From the question we are told that
The concentration of HF is [HF] = 0.6 M
The concentration of NaF is [tex][NaF ] = 0.2 M[/tex]
The Ka of HF is [tex]K_a = 6.8 *10^{-4} \[/tex]
Generally HF(Hydrogen fluoride ) is ionized as follows
[tex]HF_{(aq)} + H_2 O _{(l)} \rightleftharpoons H_3O^+ _{(aq) } + F^- _{(aq)}[/tex]
Generally NaF(Sodium fluoride ) is ionized as follows
[tex]NaF_{(aq)} + H_2 O _{(l)} \rightleftharpoons Na^+ _{(aq) } + F^- _{(aq)}[/tex]
Generally the from Henderson-Hasselbalch equation the pH of the buffer is mathematically represented as
[tex]pH = pKa + log [\frac{[NaF ]}{HF} ][/tex]
=> [tex]pH = -log(K_a) + log [\frac{[NaF ]}{HF} ][/tex]
=> [tex]pH = -log(6.8 *10^{-4}) + log [\frac{[0.2 ]}{0.6} ][/tex]
=> [tex]pH = 2.69[/tex]