MULTIPLE CHOICE QUESTION
1 answer:
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2-bromo-1-chloro-4-nitrobenzene is being synthesized in following sequence:
Step 1: Chlorination of Benzene:
This is Halogenation reaction of benzene. In this step benzene is reacted with Chlorine gas in the presence of lewis acid (i.e. FeCl₃). This results in the formation of Chlorobenzene as shown in red step below.
Step 2: Nitration of Chlorobenzene:
The chlorine atom on benzene has a ortho para directing effect. Therefore, the nitration of chlorobenzene will yield para nitro chlorobenzene as shown in blue step below.
Step 3: Bromination of 1-chloro-4-nitrobenzene:
In this step bromination is done by reacting bromine in the presence of lewis acid. The chlorine being ortho para directing in nature and nitro group being meta directing in nature will direct the incoming Br⁺ (electrophile) to the desired location. Hence, 2-bromo-1-chloro-4-nitrobenzene is synthesized in good yield.
Here is the full question
Find the pH during the titration of 20.00 mL of 0.1000 M butanoic acid, CH3CH2CH2COOH (K a = 1.54 × 10 − 5), with 0.1000 M NaOH solution after the following additions of titrant (total volume of added base given):
a) 10.00 mL
pH = <u> </u>
b) 20.10 mL
pH = <u> </u>
c) 25.00 mL
pH = <u> </u>
<u />
Answer:
pH = 4.81
pH = 10.40
pH = 12.04
Explanation:
a)
Number of moles of butanoic acid
=
= 0.002000 mol
Number of moles of NaOH added
=
= 0.001000 mol
pKa of butanoic acid = - log Ka
= - log ( 1.54 × 10⁻⁵)
= 4.81
Equation for the reaction is expressed as follows:
CH₃CH₂CH₂COOH + OH⁻ -----> CH₃CH₂COO⁻ + H₂O
The ICE Table is expressed as follows:
CH₃CH₂CH₂COOH + OH⁻ -----> CH₃CH₂COO⁻ + H₂O
Initial 0.002000 0.001000 0
Change - 0.001000 - 0.001000 + 0.001000
Equilibrium 0.001000 0 0.001000
Total Volume = (20.00 + 10.00 ) mL
= 30.00 mL = 0.03000 L
Concentration of [CH₃CH₂CH₂COOH] =
= 0.03333 M
Concentration of [CH₃CH₂COO⁻] =
= 0.03333 M
By Henderson- Hasselbalch equation
pH = pKa + log
pH = pKa + log
PH = 4.81 + log
pH = 4.81
Thus; the pH of the resulted buffer solution after 10.00 mL of NaOH was added = 4.81
b )
After the equivalence point, we all know that the pH of the solution will now definitely be determined by the excess H⁺
Number of moles of butanoic acid
=
= 0.002000 mol
Number of moles of NaOH added
=
= 0.002010 mol
Following the previous equation of reaction , The ICE Table for this process is as follows:
CH₃CH₂CH₂COOH + OH⁻ -----> CH₃CH₂COO⁻ + H₂O
Initial 0.002000 0.002010 0
Change - 0.002000 -0.002000 + 0.002000
Equilibrium 0 0.000010 0.002000
We can see here that the base is present in excess;
NOW, number of moles of base present in excess
= ( 0.002010 - 0.002000) mol
= 0.000010 mol
Total Volume = (20.00 + 20.10 ) mL
= 40.10 mL ×
= 0.04010 L
Concentration of acid [OH⁻] =
=
Using the ionic product of water:
where
=
pH = - log
pH = - log []
pH = 10.40
Thus, the pH of the solution after the equivalence point = 10.40
c)
After the equivalence point, pH of the solution is determined by the excess H⁺.
Number of moles of butanoic acid
=
= 0.002000 mol
Number of moles of NaOH added
=
= 0.002500 mol
From our chemical equation; The ICE Table can be illustrated as follows:
CH₃CH₂CH₂COOH + OH⁻ -----> CH₃CH₂COO⁻ + H₂O
Initial 0.002000 0.002500 0
Change - 0.002000 -0.002000 +0.002000
Equilibrium 0 0.000500 0.002000
Base is present in excess
Number of moles of base present in excess = [ 0.002500 - 0.002000] mol
= 0.000500 mol
Total Volume = ( 20.00 + 25.00 ) mL
= 45.00 mL
= 45.00 ×
= 0.04500 L
Concentration of acid [OH⁻] =
= 0.01111 M
Using the ionic product of water
=
= M
pH = - log
pH = - log []
pH = 12.04
Thus, the pH of the solution after the equivalence point = 12.04