wiz-icon
MyQuestionIcon
MyQuestionIcon
12
You visited us 12 times! Enjoying our articles? Unlock Full Access!
Question

Which of the following will not increase the acidic nature of B(OH)3?

A
Glycerol
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
Mannitol
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
Ethanol
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
Both (a) and (b)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C Ethanol
When certain organic polyhydroxy compounds such as glycerol, mannitol or sugars are added, B(OH)3 behaves as a strong monobasic acid.
Here, cis-diol forms a very stable complex with [B(OH)4], removing it from the solution.


Thus, the removal of one of the products shifts the equilibrium in the forward direction.
B(OH)3+2H2OB(OH)4+H3O+

Hence, it makes the B(OH)3 strong monobasic acid.
Ethanol is not a diol so it does not form complex with [B(OH)4].

THEORY
Properties of orthoboric acid
  1. White crystalline solid with soapy touch.
  2. Sparingly soluble in water but highly soluble in hot water.
  3. It has a layer structure in which planar BO3 units are joined by hydrogen bonds.
  4. Boric acid is a weak monobasic acid.
  5. It is not a protonic acid but acts as a Lewis acid by accepting electrons from a hydroxyl ion.
B(OH)3+2H2OB(OH)4+H3O+
If certain organic polyhydroxy compounds such as glycerol, mannitol, or sugars are added, B(OH)3 behaves as a strong monobasic acid.
The cis-diol forms a very stable complex with [B(OH)4], removing it from the solution. The reaction is reversible. Thus, the removal of one of the products shifts the equilibrium in the forward direction.
B(OH)3+2H2OB(OH)4+H3O+

Ethanol does not form similar complexes, but catechol, salicylic acids, and mannitol form similar complexes.



Polymeric metaborate species are formed at higher concentrations.
3B(OH)3H3O++[B3O3(OH)4]+H2O

On heating, orthoboric acid above 370K forms metaboric acid, HBO2 which on further heating yields boric oxide, B2O3.
H3BO3HBO2B2O3

Test for Borate radical:
When H3BO3 is heated with C2H5OH, the evolved gas is burned to form a green-edged flame.
H3BO3+3C2H5OHB(OC2H5)3ethyl borate+3H2O


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon