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

Answer:

pH = 11.95≈12

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

$0.1 \frac{mol NaOH}{L} * 0.030 L = 0.003 mol NaOH[tex]0.1 \frac{mol CH_3COOH}{L} * 0.025 L= 0.0025 mol CH_3COOHSecond step. Know wich substance is in excess.0.0025 mol CH_3COOH * [tex]1 mol NaOH/ 1 mol CH_3COOH$ = 0.0025 mol NaOH

0.003 mol NaOH * $1 mol CH_3COOH$/ 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

$pH= -log [H+]\\pH= -log \frac{K_w}{[OH-]} \\pH= 11.95$