Acid & Base Buffer questions

Question

Find the pH and the volume (mL) of 0.487 M HNO₃ needed to reach the equivalence point in the titration of 2.65 L of 0.0750 M pyridine (C₅H₅N, K_b = 1.7 × 10⁻⁹).

 

Answer 1

The equivalence point is the point where the moles of acid (HNO3) equals the moles of base (pyridine).

In this reaction the mole ratio of acid to base is 1:1, as shown below...

C₅H₅N + HNO₃ => C₅H5NH⁺ + NO₃^-

moles C₅H₅N present = 2.65 L x 0.0750 mols/L = 0.19875 mols C₅H₅N

moles HNO₃ needed to reach equivalence = 0.19875 mols HNO₃

Volume of HNO₃ = 0.19875 mols x 1 L / 0.487 mols = 0.408 L = 408 mls volume of HNO₃ needed

To find the pH at equivalence, we must look at the hydrolysis of C₆H₅NH⁺ that is formed.

C₆H₅NH⁺ + H₂O ==> H₃O⁺ + C₆H₅N

Ka for C₆H₅NH⁺ = 1x10^-14 / Kb = 1x10^-14 / 1.7x10^-9 = 5.88x10^-6

Ka = 5.88x10^-6 = [H₃O⁺ ] [C₆H₅N] / [C₆H₅NH⁺]

5.88x10^-6 =(x)(x) / [C₆H₅NH⁺] and [[C₆H₅NH⁺] = 0.19875 mols / 3.058 L = 0.06499 M

5.88x10^-6 =(x)(x) / 0.06499

x² = 3.892x10^-7

x = [H₃O⁺] =6.24x10^-4 M

pH = -log [H₃O⁺] = -log 6.24x10^-4 M

pH = 3.20