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Grade: 12
        
Percentage ionization of water at a certain temperature is 3.6×10^-7 . Calculate Kw and ph
one year ago

Answers : (1)

Sahani Kumar
99 Points
							
Percentage ionization = 3.6 * 10^-7
Actual ionization = (3.6 * 10^-7)/(100) = 3.6 * 10^-9
                               
                                    2(H2O)--------------->OH-  +  H3O+
initial number.            1.                            0.             0
of moles.     
Number of moles  (1 - (3.6 * 10^-9))      (3.6 * 10^-9 
at equilibrium:                                             of OH-and                                                                                H3O+)
So now Kw = [OH-] [H3O+]
                     = [3.6 * 10^-9] [3.6 * 10^-9]
               Kw=12.96 * 10^-18
Now we know that 
- pH = log[H3O+]
-pH= log[3.6 * 10^-9]
-pH =( log3.6) +(log 10^-9)
-pH = 0.556 - 9
pH = 9 - 0.556 = 8.4
Regards 
Askiitian member 
If you like this answer please encourage us by approving thus answer. 
 
one year ago
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