Question

Suppose you have 1.00 L of an aqueous buffer containing 60.0 mmol benzoic acid (pKa = 4.20) and 40.0 mmol benzoate. Calculate the pH of this buffer.

What volume of 5.00 M NaOH would be required to increase the pH to 4.93?

Answer 1

1) P^H = P^Ka + log[Phcoo^-1]/[PhCOOH]

              = 4.20+log(40/60)

               =4.02

2)mmoles PhCOOH + mmoles PhCOO- = 60.0 + 40.0 = 100.0.

During the titration, we are converting PhCOOHor to PhCOO-, so the total mmoles of PhCOOHor plus PhCOO- will always be 100.0. Set up a reaction chart.

Mmoles . . . . . .PhCOOH + OH- ==> PhCOO^-1+ H2O
Initial . . . . . . . . .60.0 . . . .x . . . . . . .40.0
Change . . . . . . .-x . . . . .-x . . . . . . . .+x
After Reaction 60.0-x . . .0 . . . . . . .40.0+x

Find the mmoles PhCOO^- / mmoles PhCOOHor ratio at pH 4.93 using the Henderson-Hasselbalch equation:

pH = pKa + log (mmoles PhCOO- / mmoles PhCOOH)
4.93 = 4.20 + log (mmoles PhCOO- / mmoles PhCOOH)
0.73 = log (mmoles PhCOO- / mmoles PhCOOH)
10^0.73 = mmoles PhCOO- / mmoles PhCOOH = 5.37
mmoles PhCOO^- = (5.37)(mmoles PhCOOH)

So let x = mmoles of PhCOO^-. Since mmoles PhCOOH + mmoles PhCOO^- = 100.0, then
mmoles PhCOOH = 100.0 - x. Put those values into the equation we just derived.

mmoles PhCOO^- = (5.37)(mmoles PhCOOH)
x = (5.37)(100.0 - x)

x = 84.30 = mmoles PhCOO^-

Note in our "ICE" chart that mmoles HCOO^- after the reaction = 40.0+x.
84.30 = 40.0+x
x = 44.3mmoles = mmoles of OH- added

mmoles OH- added = (Molarity OH-)(mL OH-)
44.3 = 5.00x
x = 8.86mL = 8.86mL 5.00 M NaOH added