Question

I really need help with c-f please

Calculate the pH of the solution after the addition of the following amounts of 0.0576 M HNO_3 to a 50.0 mL solution of 0.0750 M aziridine. The pK_a of aziridinium is 8.04. a) 0.00 mL of HNO_3 pH = b) 9.00 mL of HNO_3 pH = c) Volume of HNO_3 equal to half the equivalence point volume pH = d) 61.9 mL of HNO_3 pH = e) Volume of HNO_3 equal to the equivalence point pH = f) 70.1 mL of HNO_3 pH =

Answer 1

c)

in the half equivalence point

pKb =14- 8.04 = 5.96

then

pOH = pKb + log(BH+/B)

HB+ = B in half equivalence point so;

pOH = 5.96

pH = 14-5.96= 8.04

e)

HNO3 for equivlanece point

mmol of acid = mmol of base

Macid*Vacid = Mbase*Vbase

0.0576 * Vacid= 0.075*50

Vacid= 0.075*50/0.0576 = 65.104 mL of acid required

f)

this is mostly acidic, since there is excess H+ strong base

mmol of HNO3 = MV = 70.1*0.0576 = 4.03776 mmol of HNO3

mmol of base = MV = 0.075*50 = 3.75

mmol of hno3 excess = 4.03776 -3.75 = 0.28776 mmol of H+

V total =) 70.1 + 50 = 120.1 mL

pH = -log(h+) = -log(0.28776 /120.1) = 2.6205