Question

Calculate the pH for each of the following cases in the titration of 25.0 mL of 0.130 M pyridine, C5H5N(aq) with 0.130 M HBr(aq): a)after the addition of 16.0 mL of Hbr b) after the addition of 25.0 mL of HBr c) after the addition of 32.0 mL of HBr

Answer 1

a)

mmol of pyridine = MV = 0.13*25 = 3.25

mmol of HBr = MV = 0.13*16 = 2.08

mmol of pyridine left = 3.25-2.08 = 1.17

mmol of pyridine conjguate formed = 2.08

this is a buffer so

pOH = pKb + log(conjugate/pyridine)

pOH = 5.21 + log(2.08/1.17)

pOH = 5.45987

ph = 14-poH = 14-5.45987

pH = 8.54013

b)

this is neutralization

[pyridine conjguate]= M1V1/(V1+V2) = 25*0.13 / (25+25) = 0.065 M

BH+ + H2O <-> H#O + B

Ka = [H+][B]/[BH+]

(10^-14)/(10^-5.21) = x*x /(0.065 -x)

x = 1.03*10^-5

[H+] =  1.03*10^-5

pH = -log( 1.03*10^-5) = 4.99

c

mmol of pyridine = MV = 0.13*25 = 3.25

mmol of HBr = MV = 0.13*32 = 4.16

mmol of HBr left = 4.16 -3.25= 0.91

[H+] = mmol H+ / Vtotal = 0.91 / (25+33) = 0.0156

pH = -log(H´+)

pH = -log(0.0156) = 1.81