Question

Calculate the pH of the solution after the addition of the following amounts of 0.0640 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) 9.13 mL of HNO3 c) Volume of HNO3 equal to half the equivalence point volume d) 54.8 mL of HNO3 e) Volume of HNO3 equal to the equivalence point f) 62.2 mL of HNO3

Answer 1

The pKa of aziridinium = 8.04

a) The no. of moles of aziridine = 50 mL * 0.075 mmol/mL = 3.75 mmol

The no. of moles of HNO3 = 0 mmol

pOH = 1/2 (pKb - Log[azirdine])

i.e. pOH = 1/2 {5.96 - Log(0.075)}

i.e. pOH = 3.54

Now, pH = 14 - pOH = 14 - 3.54 = 10.46

b) The no. of moles of HNO3 = 9.13 mL * 0.064 mmol/mL = 0.58432 mmol

pH = pKa + Log([aziridine]/[aziridinium])

i.e. pH = 8.04 + Log{(3.75 - 0.58)/0.58}

i.e. pH = 8.78

c) At the half-equivalence point, [aziridine] = [aziridinium]

i.e. pH = 8.04 + Log(1)

i.e. pH = 8.04

d) The no. of moles of HNO3 = 54.8 mL * 0.064 mmol/mL = 3.51 mmol

pH = pKa + Log([aziridine]/[aziridinium])

i.e. pH = 8.04 + Log{(3.75 - 3.51)/3.51}

i.e. pH = 6.88