Respuesta :
Answer:
a)10.46
b)8.95
c)8.04
d)6.63
e)4.75
f) 2.79
Explanation:
Step 1: Data given
Molarity of HNO3 = 0.0656 M
Volume of 0.0750 M aziridine solution = 70.0 mL = 0.07 L
pKa of aziridinium is 8.04
pKb = 14 -8.04 = 5.96
⇒aziridinium has a basic character.
⇒ Az + H₂O → AzH⁺ + OH⁻
Az = aziridine
AzH⁺ = aziridinium.
a) pH of the solution after adding 0.00 mL of HNO3:
Ka of AzH⁺ = 10^-8.04 = 9.12 *10 ^-9
Kb for Az = (1 × 10^-14) / 10^-8.04 = 1.10 × 10^-6
When no HNO₃ is added, consider the dissociation of Az.
Az + H₂O → AzH⁺ + OH⁻
⇒ The initial concentration of Az is 0.0750 M and there will react X M. There will remain (0.0750 - X)M
⇒ The initial concentration of AzH+ and OH- is 0M, there will be produced X M. At the equilibrium there will be X M
Since Kb is very small, the dissociation of Az can be negligible
and [Az] at equilibrium = (0.0750 - x) M ≈ 0.0750 M
⇒ Kb = [AzH⁺] [OH⁻] / [Az]
1.10 × 10⁻⁶ = X² / 0.0750
X = 2.87 × 10⁻⁴ = [OH-]
pOH = -log[OH⁻] = -log(2.87 × 10⁻⁴) = 3.54 2
pH = 14.00 -pOH = 14.00 - 3.542 = 10.457
b) pH of the solution after adding 9.06 mL of HNO3:
⇒ Number of initial moles of Az (before addition of HNO3 = 0.0750 M * 0.07 = 0.00525 moles = 5.25 mmoles
⇒ Number of moles of HNO3 = 0.0656 M * 0.00906 L = 0.000594 = 0.594 mmoles
This will form a basic buffer in the presence of weak base (Az).
pOH = pKb + log [Conjugate acid]/[weak base]
pOH = pKb + log [AzH+]/[AZ]
When 9.06 ml of HNO₃ is added, a part of Az reacts with H⁺ to give AzH⁺. After reaction:
[Az] = (5.25 - 0.594) / (70.0 + 9.06) = 0.0589M
[AzH⁺] = 0.594 / (70.0 + 9.06) = 0.0075M
pOH = pKb + log (0.0075M/0.0589M)
pOH = 5.96 + log (0.0075M/0.0589M)
pOH = 5.055
pH = 14 -5.055 = 8.945
c) Volume of HNO3 equal to half the equivalence point volume
At half equivalence point, [conjugate acid] = [weak base]. [Az] ≈ [AzH⁺]
pOH = 5.96 + log 1
pOH = 5.96
pH = 14 - 5.96 = 8.04 (pH = pKa)
d)pH of the solution after adding 77.0 mL of HNO3
⇒ Number of initial moles of Az (before addition of HNO₃)= 5.25 mmoles
⇒ Number of moles of HNO3 = 0.0656M * 0.77L = 5.0512 mmoles
⇒ Number of moles of the limiting reactant HNO3 / total volume = 5.0512 mmoles / (147mL) = 0.0344 M
⇒ Number of moles of Az – Number of moles of HNO₃) / total volume = (5.25 - 5.0512) /(147mL) = 0.00135 M
pOH = pKb + log [Conjugate acid]/[weak base]
pOH = 5.96 + log (0.0344 M/0.00135 M)
pOH = 7.366
pH = 14 - 7.366 = 6.634
e) Volume of HNO3 equal to the equivalence point
⇒ At the equivalence point the number of moles of the base is equal to moles of acid
⇒ Volume of HNO3 needed for the equivalence point =5.25 mmol / 0.0656 M = 80.03 ml
⇒ At the equivalence point, [Az] = 0 and [AzH+] = 5.25 / (70 + 80.03) = 0.035M
As Ka is very small, the dissociation of AzH⁺ can be negligible
so [AzH+] = 0.035M
⇒ Ka = [Az] [H⁺] / [AzH⁺]
10^-8.04 = y² / 0.035
y = 1.79 * 10^-5 = [H+]
pH = -log[H+] = -log 1.79 * 10^-5 = 4.75
f)pH of the solution after adding 83.8 mL of HNO3
⇒ Number of moles of H⁺ added from HNO₃ = (0.0656 M * 83.8 mL) = 5.5 mmol
⇒ After the addition, [H⁺] = (5.5- 5.25) / (70.0 + 83.8 mL) = 0.00163 M.
As Ka is small and due to the common ion effect in the presence of H⁺, the dissociation of Az is negligible.
⇒ pH = -log[H+] = -log (0.00163 M.) = 2.79
