...pH level...?

Question

Calculate the pH of a 1.0x10 ^ -8 M HCl...

hhmm....show your solution please....thankz!....badly needing it...

Answer 1

Firstly, you must calculate the concentration of [H3O+] that come from the dissociation of HCl:
HCl + H2O ----> H3O+ + Cl-
HCl is a strong acid, so the reaction is total. In this manner, there are 1.0*10^-8 moles H3O+ that come from HCl.
Then, you must take into account the dissociation of water:
2H2O <=> H3O+ + HO-
It is known that [H3O+]*[HO-]=10^(-14), for all solutions. So the concentration of H3O+ that come from the dissociation of water is 10^(-7).
So the total concentration of H3O+ is 10^(-8) + 10^(-7)=1.1*10^(-7)
The pH is -lg(1.1*10^(-7))=6.95

Answer 2

The concentration is so low that you need to take into account the self dissociation of water. However you don't just add 10^-7. You have to treat it as the equillibrium it truly is.

So you need to find the shift of the equilibrium of the self- dissociation of water in the presence of initial [H+]= 10^-8 (since HCl is monoprotic)

.. .. .. .. .. ..H2O <=> H+ + OH-
Initial .. .. .. .. .. .. .. .. 10^-8
Dissociate .x
Produce .. .. .. .. .. .. .. x .. .. x
At equil. .. .. .. .. .. .10^-8+x .x

Kw= [H+][OH-] =>
(10^-8 + x)x =10^-14

solve the quadratic, find x and then pH=-log[H+]=-log(x+10^-8)
so you get x= 9.5125*10^-8
and pH= -log(9.5125*10^-8+(10^-8)) =6.978 =6.98

Answer 3

that is the concentration of H+ and so you just take the -log 1.0x10^8 = 8 which is the pH