Respuesta :

superman1987
we are going to use H-H equation:

PH = Pka + log[conjugate base]/[weak acid]

 when we have here the conjugate base is NaF

and the weak acid is HF.

 so, we are going to use the Ka value to get the value of Pka.

when Pka = -logKa

                  = -log7.2 X 10^-4

                  = 3.14

then we will substitution on H-H equation:

 ∴PH = 3.14 + log (0.623)/(1.41) = 2.99

Answer:

The pH of the buffer solution  is 2.78.

Explanation:

Concentration of salt = 0.623 M

Concentration of acid = 1.41 M

Dissociation constant of acid = [tex]K_a=7.20\times 10^{-4}[/tex]

The pH of the buffer solution is given by Henderson-Hasselbalch equation:

[tex]pH=pK_a+\log \frac{[salt]}{[acid]}[/tex]

[tex]pH=-\log[K_a]+\log \frac{salt}{acid}[/tex]

[tex]pH=-\log[7.20\times 10^{-4}]+\log\frac{0.623 M}{1.41 M}[/tex]

pH = 2.78

The pH of the buffer solution  is 2.78.

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