ng Spring 18-GRUE NEMUM > Titrations levtra credit) Due Friday, March 1trh Activ

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ng Spring 18-GRUE NEMUM > Titrations levtra credit) Due Friday, March 1trh Activities and Die Dates y of St. Francis . CHEM 122 3/16/2018 1,59 PM O 5 ,/83/14/2018 oes AM -Print Calculator Periodc Table Question 7 of 18 Map Sapling Learning Calculate the pH of the solution after the addition of the following amounts of 0.0557 M HNOs to a 50.0 m solution of 0.0750 M aziridine. The pKs of aziridinium is 8.04. a) 0.00 mL of HNOs d) 63.6 mL of HNO3 Number Number PH 8.04 pH b) 9.94 mL of HNO e) Volume of HNO3 equal to the equivalence point Number Number pH = pH = c) Volume of HNOj equal to haf the equivalence point volume 71.5 mL of HNO Number Number pH= PH- Hint Previous Give Up & View Solton 9 Check Answer 0 Next tl Exit to search

Explanation / Answer

Titration of weak base (aziridine) with strong acid (HNO3)

pKa of aziridinium = 8.04

pKb = 14 - 8.04 = 5.96

Kb = 1.1 x 10^-6

(a) 0 ml HNO3 added

[aziridine] = 0.0750 M

aziridine + H2O <==> aziridineH+ + OH-

let x amount has reacted

Kb = [aziridineH+][OH-]/[aziridine]

1.1 x 10^-6 = x^2/0.0750

x = [OH-] = 2.87 x 10^-4 M

pOH = -log(2.87 x 10^-4) = 3.54

pH = 14 - 3.54 = 10.46

(b) after 9.94 ml HNO3 added

initial aziridine = 0.0750 M x 50 ml = 3.75 mmol

added HNO3 = 0.0557 M x 9.94 ml = 0.554 mmol

(aziridinium) formed = 0.554 mmol

(aziridine) remained = 3.196 mmol

pH = pKa + log(aziridine/aziridinium)

     = 8.04 + log(3.196/0.554)

     = 8.80

(c) at 1/2 equivalence point

[azirindine] remained = [aziridium] formed

pH = pKa = 8.04

(d) 63.6 ml HNO3 added

initial aziridine = 0.0750 M x 50 ml = 3.75 mmol

added HNO3 = 0.0557 M x 63.6 ml = 3.542 mmol

(aziridinium) formed = 3.542 mmol

(aziridine) remained = 0.208 mmol

pH = pKa + log(aziridine/aziridinium)

     = 8.04 + log(0.208/3.542)

     = 6.81

(e) at equivalence point

volume HNO3 added = 3.75 mmol/0.0557 M = 67.325 ml

[aziridinium] formed = 3.75 mmol/(67.325 + 50) ml = 0.032 M

aziridnium + H2O <==> aziridine + H3O+

let x amount reacted

Ka = [aziridine][H3O+]/[aziridinium]

9.12 x 10^-9 = x^2/0.032

x = [H3O+] = 1.71 x 10^-5 M

pH = -log(1.71 x 10^-5) = 4.77

(f) 71.5 ml HNO3 added

excess [HNO3] = 0.0557 M x (71.5 - 67.325) ml = 0.0133 M

pH = -log(0.0133) = 1.87

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