What is the pH of the solution created by combining 0.60 mL...?

Question

What is the pH of the solution created by combining 0.60 mL of the 0.10 M NaOH(aq) with 8.00 mL of the 0.10 M HC2H3O2(aq)? What are the pH values if you take into account that the 8.00 mL of 0.10 M Acid was first diluted with 100 mL of water?

Answer 1

Moles OH- = 0.00060 L x 0.10 M = 0.000060

Moles acetic acid = 0.0080 L x 0.10 M = 0.00080

total volume = 0.0086 L

CH3COOH + OH- >> CH3COO- + H2O

moles acid = 0.00080 - 0.000060 = 0.00074

moles acetate = 0.000060

concentration acid = 0.00074 / 0.0086 = 0.0860 M

concentration acetate = 0.000060 / 0.0086 =  0.00698 M

pKa = 4.74

pH = 4.74 + log 0.00698 / 0.0860 =3.65