Incorrect Question 24 of 32 Incorrect correct Map pling Calculate the ph of the
ID: 973926 • Letter: I
Question
Incorrect Question 24 of 32 Incorrect correct Map pling Calculate the ph of the solution after the addition of the following amounts of 0.0545 M HNO3 to a 70.0 mL solution of 0.0750 Maziridine. The pKa of aziridinium is 8.04. d) 92.1 mL of HNO3 a) 0.00 mL of HNO3 Number Number pH 5.940 pH 10.46 b) 8.57 mL of HNO e) Volume of HNO3 equal to the equivalence point Number Number pH 9.08 pH 4.77 c) Volume of HNO3 equal to half the equivalence point volume f) 99.99 mL of HNO3 Number Number pH 8.04 p H Incorrect. Previous Give Up & View Solution Try Again Next Ext ExplanationExplanation / Answer
d)
millimoles of base = 70 x 0.0750 = 5.25
millimoles of HNO3 = 0.0545 x 92.1 = 5.02
B + H+ -----------------> BH+
5.25 5.02 0 --------------> I
0.23 0 5.02 -------------> E
pOH = pKb + log [BH+ / B]
pOH = 5.96 + log (5.02 / 0.23)
pOH = 7.30
pH + pOH = 14
pH = 6.70
e)
volume HNO3 eauivalenent
70 x 0.0750 = 0.0545 x V
V = 96.3 mL
at this point only salt present
salt = C = 70 x 0.0750 / (70+96.3)
= 0.03157 M
pH = 7 - 1/2 [pKb + logC]
pH = 7 - 1/2 [5.96 + log 0.03157]
pH = 4.77
f)
HNO3 millimoles = 99.99 x 0.0545 = 5.45
[H+] = 5.45 - 5.25 / (70 +99.99) = 1.17 x 10^-3 M
pH = -log[H+]
pH = 2.93