Calculate the ph of the solution after the addition of the following amounts of 0.0656 m hno3 to a 70.0 ml solution of 0.0750 m aziridine. the pka of aziridinium is 8.04.a) 0.00 ml of hno3; ph=b) 9.06 ml of hno3; ph=c) volume of hno3 equal to half the equivalence point volume; ph=d) 77.0 ml of hno3; ph=e) volume of hno3 equal to the equivalence point; ph=f) 83.8 ml of hno3; ph=

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