Question

Calculate the pH of a solution that is 0.235M benzoic acid and 0.130M sodium benzoate, a salt whose anion is the conjugate base of benzoic acid.

Answer 1

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) =
`6.20*10^(-5)`

So:

`pKa = - log\:(Ka)`

`pKa = - log\:(6.20*10^(-5))`

`pKa = 5 - log\:6.20`

`pKa = 5 - 0.79`

`\boxed{pKa = 4.21}`

Now, using the abovementioned data for the pH formula of a buffer solution or (Henderson-Hasselbalch equation), we have:

`pH = pKa + log\:([salt])/([acid])`

`pH = 4.21 + log\:(0.130)/(0.235)`

`pH = 4.21 + log\:0.55`

`pH = 4.21 + (-0.26)`

`pH = 4.21 - 0.26`

`\boxed{\boxed{pH = 3.95}}\end{array}}\qquad\checkmark`

Note:. The pH <7, then we have an acidic solution.

I Hope this helps, greetings ... DexteR! =)