Hello!
We have the following data:
ps: we apply Ka in benzoic acid to the solution.
[acid] = 0.235 M (mol/L)
[salt] = 0.130 M (mol/L)
pKa (acetic acid buffer) =?
pH of a buffer =?
Let us first find pKa of benzoic acid, knowing that Ka (benzoic acid) = [tex] 6.20*10^{-5} [/tex]
So:
[tex] pKa = - log\:(Ka) [/tex]
[tex] pKa = - log\:(6.20*10^{-5}) [/tex]
[tex] pKa = 5 - log\:6.20 [/tex]
[tex] pKa = 5 - 0.79 [/tex]
[tex] \boxed{pKa = 4.21} [/tex]
Now, using the abovementioned data for the pH formula of a buffer solution or (Henderson-Hasselbalch equation), we have:
[tex] pH = pKa + log\:\dfrac{[salt]}{[acid]} [/tex]
[tex] pH = 4.21 + log\:\dfrac{0.130}{0.235} [/tex]
[tex] pH = 4.21 + log\:0.55 [/tex]
[tex] pH = 4.21 + (-0.26) [/tex]
[tex] pH = 4.21 - 0.26 [/tex]
[tex] \boxed{\boxed{pH = 3.95}}\end{array}}\qquad\checkmark [/tex]
Note:. The pH <7, then we have an acidic solution.
I Hope this helps, greetings ... DexteR! =)
