Answer: acid dissociation constant Ka= 2.00×10^-7
Explanation:
For the reaction
HA + H20. ----> H3O+ A-
Initially: C. 0. 0
After : C-Cx. Cx. Cx
Ka= [H3O+][A-]/[HA]
Ka= Cx × Cx/C-Cx
Ka= C²X²/C(1-x)
Ka= Cx²/1-x
Where x is degree of dissociation = 0.1% = 0.001 and c is the concentration =0.2
Ka= 0.2(0.001²)/(1-0.001)
Ka= 2.00×10^-7
Therefore the dissociation constant is
2.00×10^-7
3090000000nm
since there's 1m = 1000000000nm
<u>Answer:</u> The standard potential of the cell is 0.77 V
<u>Explanation:</u>
We know that:

The substance having highest positive
reduction potential will always get reduced and will undergo reduction reaction.
The half reaction follows:
<u>Oxidation half reaction:</u> 
<u>Reduction half reaction:</u>
( × 2)
To calculate the
of the reaction, we use the equation:

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.
Putting values in above equation follows:

Hence, the standard potential of the cell is 0.77 V