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

-be a crystalline solid

- be soluble in water

-conduct electricity

hope this : )

the missing options are :

(1) lower boiling point and a lower freezing point(2) lower boiling point and a higher freezing point(3) higher boiling point and a lower freezing point(4) higher boiling point and a higher freezing point

answer : the correct answer is option (3) higher boiling point and a lower freezing point

explanation :

when a non volatile solute is added to a solvent like water, it results in elevation of boiling point.

let us find the increase in boiling point of water when cacl₂ is added to water to make a solution of 1 m

step 1 : find molality of the solution

let us assume we have 1 l of solution.

let us convert this to grams of cacl₂.

molar mass of cacl2 is 111 g/mol

grams of cacl₂ =

mass of solute = 111 g

assuming density of solution same as water which is 1g/ml we have ,

mass of solution = density * volume

mass of solution =

mass of solution = mass of solvent + mass of solute

mass of solute = 1000g - 111g

mass of solute 889 g = 0.889 kg

the molality of the solution can be calculated as

molality = 1.12 m

step 2 : find boiling point elevation.

the elevation in boiling point is calculated as

here δtb = boiling point elevation

i = vant hoff factor which is the number of ions formed by the solute .

cacl₂ dissociates as

it forms 3 ions . therefore,

i = 3

m = molality which is 1.12 m

kb is constant for a given solvent. for water it is 0.512

the boiling point of water increases by 1.7 °c

step 3: find freezing point depression.

when a non volatile solute is added to water, it results in decrease in freezing point.

the formula to find freezing point depression is

kf is constant for water and its value is 1.86

the freezing point of water decreases by 6.2 °c

therefore we can say that, the solution will have a higher boiling point and a lower freezing point