Calculate the pH for each of the cases in the titration of 25.0 mL of 0.220 M pyridine, C5H5N(aq) with 0.220 M HBr(aq) . The Kb of pyridine is 1.7×10−9.

[A]. 9.26

[B]. 5.2

[C].4.9.

[D]. 3.1

[E].1.70

Explanation:

So, we are given the following parameters:

=> 25.0 mL of 0.220 M pyridine, C5H5N(aq), 0.220 M HBr(aq) and the Kb of pyridine =  1.7×10−9.

CASE ONE:  Before the addition of HBr to pyridine.

C5H5N(aq) + H2O <> C5H5NH^+(aq) + OH^-

Kb = [C5H5NH^+][OH^-]/[C5H5N].

 1.7×10^−9. = x^2/.190 - x.

x = 1.80 × 10^-5.

Therefore, pOH = - log[OH^-] = - log [1.80 × 10^-5] = 4.74.

Thus, pH = 14 - 4.74 = 9.26.

CASE TWO:  after addition of 12.5 mL of HBr.

Number of moles of H^+ present= (The initial concentration of H^+) ×( addition of 12.5 mL of HBr) = 0.220 × 12.5 × 10^-3= 2.7 × 1O^-3 mol.

Number of moles of C5H5N present = 25 × 10^-3 × 0.220. = 5.5 × 10^-3 mol.

Therefore at equilibrium;

=> we have that 5.5 × 10^-3 mol - 2.7 × 1O^-3 mol = 2.8 × 10^-3 for C5H5N.

=> We have 2.7 × 1O^-3 mol - 2.7 × 1O^-3 mol = 0 for H^+ and 2.7 × 1O^-3 mol for C5H5NH^+.

Therefore, pH = - log (1.0 × 10^-14/ 1.7 × 10^-9) + log (2.8 × 10^-3/ 2.7 × 10^-3) = 5.23.

CASE THREE:

Number of moles of H^+ = 17 × 10^-3 × 0.220 = 3.74 ×10^-3 mol.

At equilibrium we have;

=> for C3H5N= 5.5 × 10^-3 mol - 3.74 ×10^-3 mol = 1.76 × 10^-3 and 3.74 ×10^-3 mol for C5H5NH^+.

pH = 5.23 + log (1.76 × 10^-3/3.74 × 10^-3) = 4.9.

CASE FOUR:

Number of moles of H^+ = 25 × 10^-3 × 0.220 = 5.5 × 10^-3 moles.

Thus, at equilibrium we have;

=> For C3H5N= 5.5 × 10^-3 mol - 5.5 × 10^-3 mol = 0, H^+ = 0 and C5H5NH^+ = 5.5 × 10^-3 mol.

C5H5NH^H+ + H20 > C5H5N + H3O^+.

Thus, the molarity of C5H5NH^H+ = 5.5 × 10^-3 mol/0.05 = 0.11M. At equilibrium, C5H5NH^H+ = 0.11 - x. Where C5H5 and H3O^+ are x respectively.

Therefore, ka = x^2/ 0.11 -x. (a).

Thus, we have ka = 1.0 × 10^-14/7.4 × 10^-9 = 5.8× 10^-6.

Slotting in the ka value into the equation (a) above.

x = 6.47 × 10^-7.

pH = - log { 6.47 × 10^-7} = 3.09.

CASE FIVE:

The concentration of H^+ = 1.1 × 10^-3/ 0.055L = 0.02

pH = - log { 0.02}

pH = 1.70