Calculate the pH of the solution formed when 20.0mL of 20.0 M NaOH is added to 20.0mL of 0.20M acetic acid (Ka=1.8x10^-5).

Answer 1

In 'titration', a large excess of NaOH is added to the weak acid. The titration is carried way past the end point.

The number of moles of OH- is 20E-3 x 20 = 0.4 moles

the number of moles of H+ is 20E-3 x 0.2 = 0.004 moles

The H+ ions will neutralize 0.004 moles of OH- ions leaving 0.4 - .004 = .396 moles of OH-

The volume is now 40E-3 , so the concentration of OH- is now .396/40E-3 = 9.9 molar

The pOH is -log(9.9) = .9956 and so pH = 14 - .9956 = 13.0 (with rounding)