A-Level pH, dissociation constants and buffers: worked solution

Question

Draw a sketch of how pH changes with volume as 40.00 cm3 of 0.25 mol dm⁻³ HCl(aq) is added slowly to 25.0 cm3 of 0.10 mol dm⁻³ (CH3)2CHNH2(aq). (The axes given show pH running from 0 to 14 on the vertical axis and the volume of 0.25 mol dm⁻³ HCl(aq) / cm3 running from 0 to 40 on the horizontal axis.)

This is a strong acid (HCl) added to a weak base, (CH3)2CHNH2 (isopropylamine). Moles of base = 0.0250 dm³ x 0.10 mol dm⁻³ = 2.5 x 10⁻³ mol. HCl needed for neutralisation = 2.5 x 10⁻³ mol; volume = (2.5 x 10⁻³)/0.25 = 0.0100 dm³ = 10.0 cm³. So the equivalence point is at 10.0 cm³. Starting pH (weak base, ~0.10 mol dm⁻³) is about 11-12. A gentle buffer region follows, then a sharp fall (vertical portion) at 10.0 cm³. Because the salt formed, (CH3)2CHNH3+Cl-, is that of a weak base and strong acid, the equivalence pH is acidic (about 5-6, below 7). Beyond 10 cm³, excess strong HCl brings the curve down towards pH ~1 by 40 cm³, levelling off. Sketch: curve begins near pH ~12, falls slowly through a buffer region, drops steeply at 10 cm³ passing through an equivalence pH below 7, then flattens near pH ~1.