# 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 <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?

[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

Need a fast expert's response?

Submit orderand get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

## Comments

Assignment Expert05.01.15, 21:09Dear Anik,

You're welcome. We are glad to be helpful.

If you liked our service please press like-button beside answer field. Thank you!

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

## Leave a comment