If we use the relation, pH = – log [H_{3}O^{+}], 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}

X^{2} + 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

the equation…..X2 + 10-8 x – 10-14 = 0

is giving the result 0.5*(10^-8)

it is a good answer

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you people made a good favour to all students

its good but i think it can be done in a easy way i.e.

10^-8 M HCl is a weak acid so it should be added with concentration of water thus we have

concentartion of H^+=10^-8 + 10^-7

=10^-7(1+.01)

=1.01* 10^-7

now pH= -log [H^+]

= -log [1.01*10^-7]

= -log10^-7 -log1.01

= 7-0.004

=6.996

how calculate of PH HCl 3 molar?