Question

The pH of 1 x 10-8 M HCl solution is
pls solve​

Answer 1

If we use the relation, pH = – log [H3O+], we get pH equal to 8. But this is not correct because an acidic solution cannot have pH greater than 7. It may be noted that in very dilute acidic solution, when H+ concentrations from acid and water are comparable, the concentration of H+ from water cannot be neglected.

Therefore,

[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

I hope will be right answer

Answer 2

Answer:

As the concentration of H+ from HCl is only 10^-8 M the H+ from H2O should also be considered

therefore, net H+ concentration is..

(10^-8+10^-7)

and pH is......-log(H+)=6.95