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In: Chemistry

A 25.0 mL sample of 0.125 molL−1 pyridine (Kb=1.7×10−9) is titrated with 0.100 molL−1HCl. Calculate the...

A 25.0 mL sample of 0.125 molL−1 pyridine (Kb=1.7×10−9) is titrated with 0.100 molL−1HCl.

Calculate the pH at one-half equivalence point.

Calculate the pH at 40 mL of added acid.

Calculate the pH at 50 mL of added acid.

Expert Solution

1)

millimoles of pyridine = 25.0 x 0.125 = 3.125

concentration of HCl = 0.100 M

Kb = 1.7 x 10^-9

pKb = 8.77

a) at half - equivalence point :

pOH = pKb

pOH = 8.77

pH = 5.23

b) at 40 mL of added acid.

mmoles of HCl = 40 x 0.100 = 4

C5H5N + HCl ----------------> C5H5NH+

3.125 4 0

0 0.875 3.125

here strong acid remains.

[H+] = 0.875 / 40 + 25 = 0.01346 M

pH = -log (0.01346)

pH = 1.87

c) 50 mL of added acid.

pH = 1.60

queen_honey_blossom answered 2 years ago

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