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?

pH = 11.95≈12

Explanation:

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

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

= 0.0025 mol NaOH

0.003 mol NaOH * / 1 mol NaOH = 0.003 mol CH_3COOH" alt="0.1 \frac{mol CH_3COOH}{L} * 0.025 L= 0.0025 mol CH_3COOHSecond step. Know wich substance is in excess.0.0025 mol CH_3COOH * [tex]1 mol NaOH/ 1 mol CH_3COOH" /> = 0.0025 mol NaOH

0.003 mol NaOH * / 1 mol NaOH = 0.003 mol CH_3COOH" />

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 ->

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

particle   p is an electron and particle q is a neutron