Question

Calculate the pH of the solution after the addition of the following amounts of 0.0615 M HNO3 to a 50.0 mL solution of 0.0750 M aziridine. The pKa of aziridinium is 8.04.

a) 0.00 mL of HNO3

b) 9.69 mL of HNO3

c) Volume of HNO3 equal to half the equivalence point volume

d) 57.1 mL of HNO3

e) Volume of HNO3 equal to the equivalence point

f) 65.3 mL of HNO3

Answer 1

Solution:

Given data: pKa = 8.04 and 0.0750 M aziridine

The reaction follows, HNO₃ + Aziridine -----> HA + NO₃^-

K_eq = K_a/K_b = [HA][NO₃^-] / [HNO₃][Aziridine]

K_b = [HNO₃][Aziridine]

     = (0.0615)(0.0750) = 0.0046125

pK_b = -log (0.0046125) = 2.336063 = 2.336

By using Henderson- Hasselbalch equation pH can be calculated using known pK values,

pK_a = pH - log{[Salt] / [Acid]}     here aziridinium is salt and HNO₃ is acid

8.04 = pH - log {0.0750/0.0615} = pH - 0.0860

pH = 8.04 + 0.0860 = 8.126

Part (a)

pOH = 0.05(pKb - lg(0.075)) = 4.1

14 = pH + pOH = pH + 4.1

pH = 9.9

Part(b)

at 9.69 mL Molarity [M] = {0.00969/63}/0.05 = 0.003076 M

pK_a = pH - log {[Salt] / [Acid]}

8.04 = pH - log {[0.0750] / [0.003076]}

8.04 = pH - 1.3870

pH = 9.427

Part(c)

At this point pH = pK_a = 8.04

Part(d)

at 57.1 mL Molarity [M] = {0.0571/63}/0.05 = 0.001812 M

pK_a = pH - log {[Salt] / [Acid]}

8.04 = pH - log {[0.0750] / [0.001812]}

8.04 = pH - 1.6169

pH = 9.656

Part (e)

at the equivalence point {57.1+65.3} /2 = 61.2 mL

at 61.2 mL Molarity [M] = {0.0612/63}/0.05 = 0.001942 M

pK_a = pH - log {[Salt] / [Acid]}

8.04 = pH - log {[0.0750] / [0.001942]}

8.04 = pH - 1.5868

pH = 9.6268

Part (f)

in excess of 65.3 - 61.2 = 4.1 mL

at excess 4.1 mL Molarity [M] = {0.0041/63}/0.05 = 0.001301 M

pK_a = pH - log {[Salt] / [Acid]}

8.04 = pH - log {[0.0750] / [0.001301]}

8.04 = pH - 1.7607

pH = 9.8007