Question

Calculate the pH of the solution after the addition of each of the given amounts of 0.0679 M HNO, to a 80.0 mL solution of 0.0750 M aziridine. The pK, of aziridinium is 8.04. What is the pH of the solution after the addition of 0.00 mL HNO,? pH = What is the pH of the solution after the addition of 9.75 mL HNO,? What is the pH of the solution after the addition of a volume of HNO, equal to half the equivalence point volume? pH = What is the pH of the solution after the addition of 84.8 mL HNO,? What is the pH of the solution after the addition of a volume of HNO, equal to the equivalence point? pH = What is the pH of the solution after the addition of 91.59 mL HNO,? pH =

Answer 1

concentration of aziridine = 0.00750 M

pKa= 8.04

first we need to calculate Kb from the pKa

we know

pKa = -log Ka

Ka = 10^-pKa

=10^-8.04

=9.12 X 10^-9

_Kb = Kw/ Ka

= 10^-14 / 9.12 X 10^-9

= 1.1 X 10^-6

a) pH when 0.00 mL of HNO₃ is added

it is the weak base so we need to put ICE chart

C₂H₅N + H₂O -------> C₂H₆N⁺ + OH^-

initially 0.0750 0 0

finally 0.0750-x +x +x

Kb for C₂H₅N =1.1 X 10^-6

Kb = [C₂H₅N⁺] [OH^-] / [C₂H₅N]

1.1 X 10^-6 * 0.0750 = x²

x= 2.87 X 10^-4 M

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

x = [C₂H₅N⁺]= 2.87 X 10^-4 M

pOH = -log[OH^-]

=log( 2.87 X 10^-4 M)

pOH = 3.54

pH =14- 3.54

pH =10.46

b) pH after 9.75mL of HNO₃ is added

we need to calculate moles of C₂H₅N = 0.0750 * 0.080L = 0.006 mole

moles of C₂H₅N⁺= 2.87 X 10^-4 M * 0.080L = 2.29 X10^-5 moles

when we added 9.75 mL of 0.0679 M HNO₃

So, there in comjugate base added moles and decrease the number of moles of base

moles of HNO₃=0.0679 M * 0.00975 L = 0.000662 moles

moles of C₂H₅N in excess = 0.006 - 0.000662 = 0.00534 moles

moles of C₂H₅N⁺ = 2.29 X 10^-5moles + 0.000662 = 0.000685 moles

New Molarity:-

Total volume = 80 + 9.75 ml = 89.75 mL

[C₂H₅N] = 0.00534 moles / 0.08975 L

= 0.0594 M

[C₂H₅N⁺]=0.000685 mole / 0.08975 L

= 0.00763 M

now using Henderson Hasselbalch equation

pKb = 5.96

pOH = pKb + log[C₂H₅N⁺] / [C₂H₅N]

= 5.96 + log(0.00763 / 0.0594)

=5.068

pH = 14- 5.068

pH = 8.932

c) volume of HNO₃ equal to half the equivalence point volume

we know that at half the equivalence point volume

pK = pKa

so = pH = 8.04

d) pH after 84.8 mL of HNO₃ is added

we need to calculate moles of C₂H₅N = 0.0750 * 0.080L = 0.006 mole

moles of C₂H₅N⁺= 2.87 X 10^-4 M * 0.080L = 2.29 X10^-5 moles

when we added 84.8 mL of 0.0679 M HNO₃

So, there in comjugate base added moles and decrease the number of moles of base

moles of HNO₃=0.0679 M * 0.0848 L = 0.00576 moles

moles of C₂H₅N in excess = 0.006 - 0.00576 = 0.00024 moles

moles of C₂H₅N⁺ = 2.29 X 10^-5moles + 0.00576 = 0.00578 moles

New Molarity:-

Total volume = 80 + 84.8ml = 164.8 mL

[C₂H₅N] = 0.00024 moles / 0.1648 L

= 0.001456 M

[C₂H₅N⁺]=0.00578 mole / 0.001456 L

= 0.0397 M

now using Henderson Hasselbalch equation

pKb = 5.96

pOH = pKb + log[C₂H₅N⁺] / [C₂H₅N]

= 5.96 + log(0.0397 / 0.01456)

=6.396

pH = 14-6.396

pH = 7.604

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