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:

$pH = pKa + log\:\dfrac{[salt]}{[acid]}$

$pH = 4.21 + log\:\dfrac{0.130}{0.235}$

$pH = 4.21 + log\:0.55$

$pH = 4.21 + (-0.26)$

$pH = 4.21 - 0.26$

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

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

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