Answer:
pH = 11.95≈12
Explanation:
Remember the reaction among aqueous acetic acid ([tex]CH_3COOH[/tex]) and aqueous sodium hydroxide (NaOH)
[tex]CH_3COOH + NaOH -> CH_3COONa + H_2O[/tex]
First step. Need to know how much moles of the substances are present
[tex]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_3COOH
Second step. Know wich substance is in excess.
0.0025 mol CH_3COOH * [tex]1 mol NaOH/ 1 mol CH_3COOH[/tex] = 0.0025 mol NaOH
0.003 mol NaOH * [tex]1 mol CH_3COOH[/tex]/ 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 -> [tex]Na^{+} + OH^{-}[/tex]
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
[tex]pH= -log [H+]\\pH= -log \frac{K_w}{[OH-]} \\pH= 11.95[/tex]
