Question

Titration Lab: Titrated an unknown acid (HA) with NaOH

Using volume of base at the equivalnce point, its molarity, and the fact that you used 25.0 mL of acid, calculate the concentration of the unknown acid.

HA + NaOH --> H2O + A-

Determined that:

volume of base at eq. pt = 23.03 mL

molarity of base = 0.0239

volume of acid used = 25 ml

pka of acid = 4.43

I just need help finding the concentration of the acid. I think you have to use an ice tabel and then the henderson hesselbalch equation...

Answer 1

The acid is monoprotic, i.e, contains only one ionizable proton. The balanced chemical equation for the acid-base neutralization is given as

HA (aq) + NaOH (aq) -------> NaA (aq) + H₂O (l)

At the equivalence point,

moles NaOH = moles HA.

Find the moles of NaOH required to complete the titration from the volume and molarity of NaOH.

Mole(s) NaOH = (volume of NaOH at the equivalence point)*(molarity of NaOH) = (23.03 mL)*(1 L/1000 mL)*(0.0239 M)*(1 mol/L/1 M) = 5.50417*10^-4 mole.

Mole(s) of HA neutralized = mole(s) of NaOH required = 5.50417*10^-4 mole.

Molarity of HA = (moles of HA)/(volume of HA taken) = (5.50417*10^-4 mole)/[(25.00 mL)*(1 L/1000 mL)] = 0.02201668 mol/L ≈ 0.0220 M (ans).