Respuesta :

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]

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