Respuesta :

Answer:

[Acetic acid] = 0.07 M

[Acetate] = 0.13 M

Explanation:

pH of buffer = 5

pKa of acetic acid = 4.76

[tex]pH=p_{Ka} + log\frac{[Salt]}{[Acid]}[/tex]

Now using Henderson-Hasselbalch equation

[tex]5=4.76 + log\frac{[Acetate]}{[Acetic\;acid]}[/tex]

[tex]log\frac{[Acetate]}{[Acetic\;acid]} = 0.24[/tex]

[tex]\frac{[Acetate]}{[Acetic\;acid]} = 1.74[/tex]  ....... (1)

It is given that,

[Acetate] + [Acetic acid] = 0.2 M     ....... (2)

Now solving both the above equations

[Acetate] = 1.74[Acetic acid]

Substitute the concentration of acetate ion in equation (2)

1.74[Acetic acid] + [Acetic acid] = 0.2 M

[Acetic acid] = 0.2/2.74 = 0.07 M

[Acetate] = 0.2 - 0.07 = 0.13 M

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