Incorrect Question 3 of 14 Map Sapling Learning Calculate the pH of the solution
ID: 508279 • Letter: I
Question
Incorrect Question 3 of 14 Map Sapling Learning 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 Maziridine. The pKa of aziridinium is 8.04. d) 57.0 mL of HNO3. a) 0.00 mL of HNO3 Number Number pH 9.45 pH 10.46 b) 5.03 mL of HNO e) Volume of HNO3 equal to the equivalence point Number Number pH 8.12 pH c) Volume of HNO3 equal to half 65.0 mL of HNO3 the equivalence point volume Number Number pH 8.04 pH Incorrect. A Previous Try Again Next ExitExplanation / Answer
pKa = 8.04
pKb = 14 - 8.04 = 5.96
millimoles of azridinium = 50 x 0.0750 = 3.75
b) after the addition of 5.03 mL HNO3
millimoles of HNO3 = 5.03 x0.0615 = 0.3093
B + HNO3 -----------------> BH+
3.75 0.3093 0
3.44 0 0.3093
pOH = pKb + log [salt / base]
= 5.96 + log [0.3093 / 3.44]
= 4.91
pH = 9.09
c) after the addition of 57 mL HNO3
millimoles of acid = 57 x 0.0615 = 3.5055
B + HNO3 ----------------------> BH+
3.75 3.5055 0
0.2445 0 3.5055
pOH = pKb + log 3.5055 /0.2445)
pOH = 7.12
pH = 6.88
d) after the addition of 25 mL HNO3
millimoles of HNO3 = millimoles of base
0.0615 x V = 3.75
V = 60.98 mL
it is equivalence point only salt is formed
salt millimoles = 3.75
salt concentration = millimoles / total volume = 3.75 / (50 + 60.98) = 0.0338 M
salt is from strong acid weak base so pH <7
pH = 7 -1/2 [pKb + logC]
pH = 7 - 1/2 [5.96 + log 0.0338]
pH = 4.76
f)
millimoles of HNO3 = 65 x 0.0615 = 3.9975
[H+] = 3.9975 - 3.75 / 50 + 65 = 2.15 x 10^-3
pH = 2.67