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?

1 Answer
May 6, 2018

Consider,

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

puu.sh
*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#