Question

A 100.0-mL aliquot of 0.200 M diprotic acid H₂A (pK₁ = 4.00, pK₂= 8.00) was titrated with 1.00 M NaOH. Find the pH after 11.75 mL of NaOH have been added.

Neglect activity coefficients for this problem.

Answer 1

moles diprotic acid = 100 mL x 10^-3 L x 0.200 M = 0.02 mol

moles base = 11.75 mL x 10^-3 L x 1.00 M = 0.01175 mol

R.....................H₂A...... + ......OH-...........<==> HA^-......+ ........H₂O

I.......................0.02.............0.01175..................0..........................

C......................-0.01175......-0.01175................0.01175

E......................0.00825..............0....................0.01175

[HA^-] = 0.01175

[H₂A] = 0.00825

since all the NaOH has been consumed so there is not enough OH- ions to further deprotonate HA-, so discard pk2

This is now a buffer; pH = pKa1 + log[HA-]/H₂A] = 4.00 + log(0.01175 / 0.00825) = 4.15

pH = 4.15

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