Buffer Solution Chemistry Math Question 4

Question

When 440.0 mL of 0.45 M benzoic acid is titrated with 0.3 M NaOH, what is the pH of the solution after 190.0 mL of base has been added? The pK_a for benzoic acid is 4.19.

Answer 1

Benzoic acid = C₆H₅CO₂H

The reaction is as follows:

C₆H₅CO₂H + NaOH ==> C₆H₅CO₂Na + H2O

We can set up an ICE table to see what happens to the moles during the reaction:

moles acid = 0.440 L x 0.45 mol/L = 0.198 moles

moles base = 0.190 L x 0.3 mol/L = 0.057 moles

C₆H₅CO₂H + NaOH ==> C₆H₅CO₂Na + H₂O

0.198.............0.057.................0..............................Initial

-0.057------------0.057..............+0.057.....................Change

0.141..............0......................0.141........................Equilibrium (NOTE: moles acid = moles salt, so pH = pKa)

Total volume after reaction = 440 ml + 190 ml = 630 ml = 0.630 L

Final concentrations:

[salt] = 0.141 moles/0.630 L = 0.224 M

[acid] = 0.141 moles/0.630 L = 0.224

NOTE: concentration of acid = concentration of salt and thus pH = pKa

pH = 419

To see why this is so, look at the Henderson Hasselbalch equation;

pH = pKa + log [salt]/[acid]

pH = pKa = log 1

pH = pKa