Answer:
[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
