Question

Calculate the pH of the solution after the addition of each of the given amounts of 0.0541 M HNO, to a 60.0 mL solution of 0.0750 M aziridine. The pK, of aziridinium is 8.04. What is the pH of the solution after the addition of 0.00 mL HNO,? pH = What is the pH of the solution after the addition of 8.32 mL HNO,? pH = What is the pH of the solution after the addition of a volume of HNO, equal to half the equivalence point volume? pH = What is the pH of the solution after the addition of 79.2 mL HNO,? pH = What is the pH of the solution after the addition of a volume of HNO, equal to the equivalence point? PH= What is the pH of the solution after the addition of 87.12 mL HNO,? pH =

Answer 1

(1.) What is the pH of the solution after the addition of 0.00 mL HNO₃? : pH = 10.46

(2.) What is the pH of the solution after the addition of 8.32 mL HNO₃? : pH = 8.99

(3.) What is the pH of the solution after the addition of a volume of HNO₃ equal to half the equivalence point volume? : pH = 8.04

(4.) What is the pH of the solution after the addition of 79.2 mL HNO₃? : pH = 6.74

(5.) What is the pH of the solution after the addition of a volume of HNO₃ equal to the equivalence point volume? : pH = 4.77

(6.) What is the pH of the solution after the addition of 87.12 mL HNO₃? : pH = 2.84

Explanation

(1.) pK_a of aziridinium = 8.04

pK_b of aziridine = pK_w - pK_a

pK_b of aziridine = 14 - 8.04

pK_b of aziridine = 5.96

K_b = 10^-pK_b

K_b = 10^-5.96

K_b = 1.1 x 10^-6

initial concentration of aziridine = 0.0750 M

ICE table	aziridine	H2O	$\rightleftharpoons$	aziridinium	OH-
Initial conc.	0.0750 M	-		0	0
Change	-x	-		+x	+x
Equilibrium conc.	0.0750 M - x	-		+x	+x

K_b = [aziridinium]_eq[OH^-]_eq / [aziridine]_eq

1.1 x 10^-6 = [(x) * (x)] / (0.0750 M - x)

Solving for x, x = 2.86 x 10^-4 M

[OH^-] = x = 2.86 x 10^-4 M

pOH = -log[OH^-]

pOH = -log(2.86 x 10^-4 M)

pOH = 3.54

pH = 14 - pOH

pH = 14 - 3.54

pH =