Question

please help determine the circled area from the data recorded during titration

Exp 7 TRIAL 1 ation Results TRIAL 1 NaOH volume at equivalencepH 3.02 volume OmL pH 4.41 vol. 3mL point NaOH volume at half H 3.14 volume .5mL pH 4.50 vol. 3.5mL equivalence point pH at the equivalence point pH 3.73 volume 1mL pH at half equivalence pointpH 3.96 volume 1.5mL pKa of acetic acid Ka of acetic acid pH 4.63 pH 4.74 pH 4.88 pH 5.14 vol. 4mL vol. 4.5mL vol. 5ml vol. 5.5mL pH 4.06 volume 2mL pH 4.20 volume 2.5mL TRIAL 1 pH 5.31 pH 5.55 pH 6.34 pH 7.18 TRIAL 1 pH 11.36 vol. 7.4mL pH 11.44 vol. 7.5mL pH 11.59vol. 8.1mL pH 11.72 vol. 8.6mL vol. 6mL vol. 6.5mL vol. 7mL vol. 7.1mL HALF EQUIVALENCE POINT H 10.64 vol. 7.2mL pH 10.77 vol. 7.3mL

Answer 1

Let's me correct one thing it is equivalent point when 7.1 ml of NaOH is added and pH is 7.18 not equivalent point.

NaOH volume at equivalent point = 7.1 ml

NaOH volume at half equivalent point = 7.1/2 = 3.55 ml

pH at equivalent point = 7.18

pH at half equivalent point = 4.5

pKa of acetic acid = 4.5

Ka = 10^-4.5 = 3.16 × 10^-5