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monitta
3 years ago
5

Aldehydes and ketones can be halogenated at their α-position by reaction with Cl2, Br2, or I2, under acidic conditions. Using Br

2 under acidic conditions, an intermediate enol is formed which adds bromine at the α-position. The reaction stops after the addition of one bromine because the electron-withdrawing halogen decreases the basicity of the carbonyl oxygen, making the protonation less favorable. Draw curved arrows to show the movement of electrons in this step of the mechanism.

Chemistry
1 answer:
notka56 [123]3 years ago
5 0

Answer:

Look at the picture

Explanation:

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NH3(g) will take the shape of and completely fill a closed 100.0 milliliter container. 
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Calculate the ph of a 0.60 m h2so3, solution that has the stepwise dissociation constants ka1 = 1.5 × 10-2 and 1.82 1.06 1.02 2.
Vsevolod [243]
Missing in your question Ka2 =6.3x10^-8
From this reaction:
 H2SO3 + H2O ↔ H3O+  + HSO3-
by using the ICE table :
                H2SO3     ↔    H3O     +    HSO3- 
intial         0.6                     0                  0
change     -X                      +X                +X
Equ         (0.6-X)                  X                   X

when Ka1 = [H3O+][HSO3-]/[H2SO3]
So by substitution:
1.5X10^-2 = (X*X) / (0.6-X) by solving this equation for X
∴ X = 0.088
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by using the ICE table 2:
                 HSO3-     ↔   H3O     +     SO3-
initial        0.088              0.088              0
change    -X                      +X                   +X
Equ         (0.088-X)          (0.088+X)          X

Ka2= [H3O+] [SO3-] / [HSO3-]
we can assume [HSO3-] =  0.088 as the value of Ka2 is very small
 6.3x10^-8 = (0.088+X)*X / 0.088
X^2 +0.088 X - 5.5x10^-9= 0 by solving this equation for X
∴X= 6.3x10^-8
∴[H3O+] = 0.088 + 6.3x10^-8
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∴PH= -㏒[H3O+]
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Answer:

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