Question

50.0 mL of 0.10 M acetic acid is titrated with 10.0 mL of 0.10 M NaOH. What's the pH of the solution?

Answer 1

Consider,

`HAc + OH^- rightleftharpoons H_2O + Ac^(-)`

*data given in moles

We assume the hydroxide ion is a strong base and will "totally" react with one equivalent of protons from acetic acid molecules.

Now, recall,

`"p"H = "p"K_"a" + log(([A^-])/([HA]))`

The concentrations of interest at equilibrium are,

`[HA] = [HAc]_"eq" approx 6.7*10^-2"M"`, and

`[A^-] = [Ac^-]_"eq" approx 1.7*10^-2"M"`

The `K_"a"` of acetic acid is `1.8*10^-5`.

Hence,

`"p"H = "p"K_"a" + log(([Ac^-])/([HAc])) approx 4.15`