Question

1) For 10mL of .50 M Morphine (pK_b = 6.05) give the pH at the various points indicated when titrated with 0.30 M HNO₃. K_w = 1 x10^-14

a. before any HNO₃ has been added

b. At 1/4 equivalence point

c. at 1/2 the equivalence point

d. at 3mL past the equivalence point

Answer 1

Morphine millimoles = 10 x 0.5 = 5

pKb = 6.05

Kb = 8.91 x 10^-7

a. before any HNO3 has been added

pOH = 1/2 (pKb + log C)

= 1/2 (6.05 + log 0.5)

= 2.87

pH = 11.13

c. at 1/2 the equivalence point

at half equivalence point

pOH = pKb = 6.05

pH = 7.95

d) at 3mL past the equivalence point

millimoles of base = millimoles of acid

5 = 0.3 x V

V = 16.67 mL

volume of HNO3 = 16.67 mL

3 mL past means = 16.67 + 3 = 19.67 mL

millimoles of HNO3 = 19.67 x 0.3 = 5.9

morphine + HNO3 -----------------> salt

5 5.9 0

0 0.9 5

here strong acid remained. so

[H+] = 0.9 / (19.67 + 10) = 0.0303 M

pH = -log [H+] = -log (0.0303)

= 1.52

pH = 1.52