Question

25 mL of 0.10 M aqueous acetic acid is titrated with 0.10 M NaOH(aq). What is the pH after 30 mL of NaOH have been added?

Answer 1

pH = 11.95≈12

Step-by-step explanation:

Remember the reaction among aqueous acetic acid (
`CH_3COOH`) and aqueous sodium hydroxide (NaOH)

`CH_3COOH + NaOH ->  CH_3COONa + H_2O`

First step. Need to know how much moles of the substances are present

= 0.0025 mol NaOH

0.003 mol NaOH *
`1 mol CH_3COOH`/ 1 mol NaOH = 0.003 mol CH_3COOH[/tex]

NaOH is in excess. Now, how much?

0.003 mole NaOH - 0.0025 mole NaOH = 0.0005 mole NaOH

Then, that amount in excess would be responsable for the pH.

Third step. Know the pH

Remember that pH= -log[H+]

According to the dissociation of water equilibrium

Kw=[H+]*[OH-]= 10^(-14)

The dissociation of NaOH is

NaOH ->
`Na^(+) + OH^(-)`

Now, concentration of OH^{-}[/tex] would be given for the excess of NaOH.

[OH-]= 0.0005 mole / 0.055 L = 0.00909 M

Careful: we have to use the total volumen

Les us to calculate pH

`pH= -log [H+]\\pH= -log (K_w)/([OH-]) \\pH= 11.95`