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 <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?

Expert's answer

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Assignment Expert05.01.15, 21:09Dear Anik,

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Anik03.01.15, 18:25Thanx A LOT.....................Sir

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