Please show ALL work. If written please show with neat handwriting. I will rate! Thank You! Sketch a realistic titration curve for the following situation. For last name starting: A-L Titrate 20.00 mL of 0.300 M morpholine with 0.200 M HCl M-O Titrate 10.00 mL of 0.300 M sodium 4-nitrophenolate with 0.100 M HCI P-Z Titrate 30.00 mL of 0.100 M sodium cyanide with 0.200 M HOI You should show proof of some specific points in the titration curve. i) initial pH ii) pH at 9.00 mL addition i) pH at the equiv. point iv) pH at 2.00 mL past the equiv. pt. Solution Titration Morpholine with HCl i) initial pH [morpholine] = 0.3 M morpholine + H2O <==> morpholineH+ + OH- let x amount hydrolyzed Kb = [morpholineH+][OH-]/[morpholine] 2.5 x 10^-6 = x^2/0.3 x = [OH-] = 8.7 x 10^-4 M [H+] = 1 x 10^-14/8.7 x 10^-4 = 1.15 x 10^-11 M pH = -log[H+] = 10.94 ii) 9.0 ml HCl added moles morpholine = 0.3 M x 20 ml = 6 mmol moles HCl = 0.2 M x 9 ml = 1.8 mmol morpholineH+ formed = 1.8 mmol morpholine remained = 6-1.8 = 4.2 mmol pH = pKa + log(morpholine/morpholneH+) = 8.4 + log(4.2/1.8) = 8.77 iii) pH at equivalence point volume HCl added = 6 mmol/0.2 M = 30 ml [morpholineH+] formed = 6 mmo/50 ml = 0.12 M morpholineH+ + H2O <==> morpholine + H3O+ let x amount hydrolyzed Ka = [morpholine][H3O+]/[morpholineH+] 4 x 10^-9 = x^2/0.12 x = [H3O+] = 2.2 x 10^-5 M pH = -log[H3O+] = 4.66 iv) pH at 2 ml past equivalence point excess [HCl] = [H+] = 0.2 M x 2 ml/52 ml = 0.008 M pH = -log(0.008) = 2.11 v) pH at half-equivalence point [morpholineH+] formed = [morpholine] remained pH = pKa = 8.4 Plot, volume HCl on x-axis and pH on y-axis .

1. Please show ALL work. If written please show with neat handwriting. I will rate! Thank You!
Sketch a realistic titration curve for the following situation. For last name starting: A-L Titrate
20.00 mL of 0.300 M morpholine with 0.200 M HCl M-O Titrate 10.00 mL of 0.300 M sodium
4-nitrophenolate with 0.100 M HCI P-Z Titrate 30.00 mL of 0.100 M sodium cyanide with 0.200
M HOI You should show proof of some specific points in the titration curve. i) initial pH ii) pH
at 9.00 mL addition i) pH at the equiv. point iv) pH at 2.00 mL past the equiv. pt.
Solution
Titration
Morpholine with HCl
i) initial pH
[morpholine] = 0.3 M
morpholine + H2O <==> morpholineH+ + OH-
let x amount hydrolyzed
Kb = [morpholineH+][OH-]/[morpholine]
2.5 x 10^-6 = x^2/0.3
x = [OH-] = 8.7 x 10^-4 M
[H+] = 1 x 10^-14/8.7 x 10^-4 = 1.15 x 10^-11 M
pH = -log[H+] = 10.94
ii) 9.0 ml HCl added
moles morpholine = 0.3 M x 20 ml = 6 mmol

2. moles HCl = 0.2 M x 9 ml = 1.8 mmol
morpholineH+ formed = 1.8 mmol
morpholine remained = 6-1.8 = 4.2 mmol
pH = pKa + log(morpholine/morpholneH+)
= 8.4 + log(4.2/1.8)
= 8.77
iii) pH at equivalence point
volume HCl added = 6 mmol/0.2 M = 30 ml
[morpholineH+] formed = 6 mmo/50 ml = 0.12 M
morpholineH+ + H2O <==> morpholine + H3O+
let x amount hydrolyzed
Ka = [morpholine][H3O+]/[morpholineH+]
4 x 10^-9 = x^2/0.12
x = [H3O+] = 2.2 x 10^-5 M
pH = -log[H3O+] = 4.66
iv) pH at 2 ml past equivalence point
excess [HCl] = [H+] = 0.2 M x 2 ml/52 ml = 0.008 M
pH = -log(0.008) = 2.11
v) pH at half-equivalence point
[morpholineH+] formed = [morpholine] remained
pH = pKa = 8.4
Plot,
volume HCl on x-axis and pH on y-axis