Question

Consider the titration of the weak acid HA with NaOH. At what fraction of Ve does pH = pKa -1? At what fraction of Ve does pH = pKa +1? Use these two points, plus Vb = 0, 1/2Ve, Ve, and 1.2Ve to sketch the titration curve for the reaction of 100 mL of 0.100M anilinium bromide ("aminobenzene x HBr") with 0.100M NaOH.

Answer 1

At EP Vb = 100 mLxo.1M/0.1M = 100 mL

Before EP the solution contains a buffer A^-/HA

pH = pKa +log ([A-]/[HA])

[A-]/[HA] = 10^pH-pKa

If pH = pKa-1 then   [A-]/[HA] = 0.1     ([A-] = 9%, [HA] = 91% )

In this moment of the titration Vb = 0.09Ve    (you can approximate to 0.1)

Vb = 9 mL

If pH = pKa+1 then   [A-]/[HA] = 10      ([A-] = 91%, [HA] = 9% )

In this moment of the titration Vb = 0.91Ve   (you can approximate to 0.9)

Vb = 91 mL

…………………..

Anilinium cation has a pKa=4.63

For Vb= 0

pH = 0.5pKa -0.5logC_acid = 2.3 + 0.5 = 2.8

At EP Vb = 100 mLxo.1M/0.1M = 100 mL and the solution contains only a weak base (aniline), its concentration being 0.1M x100 mL/200mL = 0.05M

pOH = 0.5pKb -0.5 logCbase = 4.7 +0.65 = 5.3

pH = 14-pOH = 8.7

At Vb = 120 mL the solution contains the excess of NaOH

[HO-] = 20 mL x 0.1 M/220 mL = 0.01 M pOH= 2.0

pH= 12.0

For Vb = 9mL , pH = 3.63

For Vb= 50 mL , pH = 4.63

For Vb = 91 mL, pH = 5.63

Plot these values.