A 25.0 mL solution of quinine was titrated with 1.00 M hydrochloric acid, HCl. It was found that the solution contained 0.125 moles of quinine. What was the pH of the solution after 50.00 mL of the HCl solution were added

pH = 9.08

Explanation:

Quinine, C₂₀H₂₄O₂N₂, Q, is a weak base that, in water, has as equilibrium:

Q + H₂O ⇄ QH⁺ + OH⁻

Where pKb is 5.10

Using H-H equation for weak bases:

pOH = pKb + log₁₀ [QH⁺] / [Q]

The reaction of quinine with HCl is:

Q + HCl → QH⁺ + Cl⁻

Initial moles of quinine are 0.125 moles and moles added of HCl are:

0.05000L × (1.00mol / L) = 0.05000moles.

That means after the addition of 50.00mL of the HCl solution, moles of Q and QH⁺ are:

Q = 0.125mol - 0.050mol = 0.075 moles

QH⁺ = 0.050 moles

Replacing in H-H equation:

pOH = 5.10 + log₁₀ [0.050] / [0.075]

pOH = 4.92

As pH = 14 - pOJ