<h3><u>Answer;</u></h3>
0.5 M HBr, pOH = 13.5 ; Has the lowest pH
<h3><u>Explanation;</u></h3>
From the question;
pH = -Log [OH]
or pH = 14 - pOH
Therefore;
For 0.5 M HBr
[H+] = 0.5 M
pH = - Log [0.5]
= 0.30
For; pOH = 13.5
pH = 14 - pOH
= 14 -13.5
= 0.5
For; 0.05 M HCl
pH = - log [H+]
[H+] = 0.05
pH = - Log [0.05]
= 1.30
For; pOH = 12.7
pH = 14 -pOH
= 14 -12.7
= 1.30
For; 0.005 M KOH,
pOH = - log [OH]
[OH-] = 0.005
pOH = - Log 0.005
= 2.30
pH = 14 - 2.30
= 11.7
For; pOH = 2.3
pH = 14 -pOH
= 14- 2.3
= 11.7
Given :
Length , l = 59.8 m.
Breadth , b = 26.6 m.
Depth , d = 3.7 ft .
Density of water , 
.
To Find :
Mass of water in pool .
Solution :
First we will covert depth into m from ft .
For ,
So , volume of pool is :
We know , density is given by :
So , 
Putting given values in above equation :
Hence , this is the required solution.
Answer:
The thermodynamic parameter which is of significance in this case is the 'Reduction Potential' for molecular bromine which is ~ +1.1 v vs N.H.E. In other words, it is a strong oxidizing agent. The bromine will oxidize sulfur compounds in which the valence of sulfur is lower than six to sulfate.
There are many possible reactions. Here is one possible example:
Na2 S2O3 + 4Br2 + 5 H2O = 2NaHSO4 + 8 HBr
