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Here is the ques that I tried to cut/paste as a picture in above post

34. Consider the following equilibria.
HNO2 + BrO- <--> HBrO + NO2- Keq = 2.4 x 105
H2SiO3 + Br0- <--> HBrO + HSiO3- keq = 9.5 x 10-2
HIO4 + NO2 <--> HNO2 + IO4 Keq = 45
The strongest base is:
A. HSiO3 B. N02 C. BrO D. I04
(Original post by hydronium)
Here is the ques that I tried to cut/paste as a picture in above post

34. Consider the following equilibria.
HNO2 + BrO- <--> HBrO + NO2- Keq = 2.4 x 105
H2SiO3 + Br0- <--> HBrO + HSiO3- keq = 9.5 x 10-2
HIO4 + NO2 <--> HNO2 + IO4 Keq = 45
The strongest base is:
A. HSiO3 B. N02 C. BrO D. I04
In each of the equations the species that gains a hydrogen ion is behaving as a Bronsted Lowry base.

The larger the equilibrium constant the more the equilibrium lies to the RHS and the stronger the base.

However, then you have to take into account the relative strengths of the acids involved.

In equation 1, the equilibrium lies well to the RHS which means that BrO- can remove a hydrogen ion from HNO2 but NO2 cannot remove a hydrogen ion from HBrO. Hence BrO- is a stronger base than NO2.

Apply similar logic to the oter equations.

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Last updated: July 2, 2012
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