Question

What is the pH of the resulting solution if 30.00 mL of 0.10 M acetic acid is added to 30.00 mL of 0.10 M NaOH?

K_a = 1.8x10^-5 for CH₃CO₂H.

Answer 1

10.00 mL x 0.10 M = 1.0 millimole NaOH
30.00 mL x 0.10 M = 3.0 millimole HOAc

HOAc + NaOH --> NaOAc + H2O

NaOAc = 1.0 millimole
HOAc = 2.0 millimole (3.0 - 1.0 = 2.0)

pH = pKa + log(AcO^-1/HOAc)
pKa = -log(1.8x10^-5) = 4.74
pH = 4.74 + log(1/3) = 4.74 - 0.48 = 4.26
pH = 4.26 (4.3; 2 s.f.)

Answer 2

millimoles of HCl =30.00 mL* 0.10 M = 3.0 millimoles ---- similarly millimoles of NaOH =30.00 mL* 0.10 M = 3.0 millimoles ----- Thus there is excess of HCl (because one mole of HCl reacts with one mole of NaOH) ---- thus excess HCl = 3.0 millimoles - 1.0 millimoles = 2.0 millimoles =2.0 X 10^-3 mole ---- Total volume = 30.00 mL + 30.00 mL = 60.00 mL =0.060 mL ---- thus final molar concentration of HCl = 2.0 X 10^-3 / 0.060 = 0.05 ---- pH = - log [H+] = - log 0.033 = 1.4775