Answer to Question #50178 in Physical Chemistry for Anik

Question #50178
For the ques :- PH of 10 * 10^(-8) of HCL.. the solution given is [H+] total = [H+] acid + [H+] water Since HCl is a strong acid and is completely ionized [H+] HCl = 1.0 x 10-8 The concentration of H+ from ionization is equal to the [OH-] from water, [H+] H2O = [OH-] H2O = x (say) [H+] total = 1.0 x 10-8 + x But [H+] [OH-] = 1.0 x 10-14 (1.0 x 10-8 + x) (x) = 1.0 x 10-14 X2 + 10-8 x – 10-14 = 0 Solving for x, we get x = 9.5 x 10-8 Therefore, [H+] = 1.0 x 10-8 + 9.5 x 10-8 = 10.5 x 10-8 = 1.05 x 10-7 pH = – log [H+] = – log (1.05 x 10-7) = 6.98 so, x= [OH-] =9.5 X 10-8, so, POH is 7.02 But If it is so what would be the case for 10*-7 M of HCL, x = [OH-] would be 1.6 X 10-7 and POH &lt;7, BUT HOW IT IS POSSIBLE as it cant be less than 7 and PH-POH=PH would be more than 7 . HOW it is possible?

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Assignment Expert
05.01.15, 21:09

Dear Anik,
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Anik
03.01.15, 18:25

Thanx A LOT.....................Sir