Question

a) Calculate the pH for a solution containing 3.47M C₅H₅N and 2.38M C₅H₅NHBr. K_b for C₅H₅N is 1.40x10^-9

b) Calculate the pH after 125.0 mL of 3.67M NaOH is added to 575.0 mL of the solution in part a) above.

Answer 1

a) mixture of C5H5N and C5H5NHBr act as basic buffer

pOH = pKb + log [C5H5NHBr] / [C5H5N]

pKb = - log Kb = - log [1.40 x 10^-9]

pKb = 8.85

pOH = 8.85 + log [2.38] / [3.47]

pOH = 8.69

pH = 14 - 8.69

pH = 5.31

b) if we add NaOH

initial millimoles of C5H5N = 575 x 3.47 = 1995.25

initial millimoles of C5H5NHBr = 575 x 2.38 = 1368.5

millimoles of NaOH added = 125 x 3.67 = 44.04

after NaOH added

[C5H5N] = [1995.25 + 44.04 / 700] = 2.91 M

[C5H5NHBr] = [1368-44.04 / 700] = 1.89M

pOH = 8.85 + log [1.89] / [2.91]

pOH = 9.04

pH = 14 - 9.04

pH = 4.96