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