Answer:
4.76
Explanation:
In this case, we have to start with the buffer system:
[tex]CH_3COOH~->~CH_3COO^-~+~H^+[/tex]
We have an acid ([tex]CH_3COOH[/tex]) and a base ([tex]CH_3COO^-[/tex]). Therefore we can write the henderson-hasselbach reaction:
[tex]pH~=~pKa+Log\frac{[CH_3COO^-]}{[CH_3COOH]}[/tex]
If we want to calculate the pH, we have to calculate the pKa:
[tex]pH=-Log~Ka=4.76[/tex]
According to the problem, we have the same concentration for the acid and the base 0.1M. Therefore:
[tex][CH_3COO^-]=[CH_3COOH][/tex]
If we divide:
[tex]\frac{[CH_3COO^-]}{[CH_3COOH]}~=~1[/tex]
If we do the Log of 1:
[tex]Log~1=~zero[/tex]
So:
[tex]pH~=~pKa[/tex]
With this in mind, the pH is 4.76.
I hope it helps!
