Calculate the pH of a 3.26 x 10-8 M solution of HBr.?

Answer 1

The concentration is so low that you need to take into account the self-dissociation of water. However since the latter is an equilibrium, you can't just add 10^-7 to the H+ coming from the acid.

There are 2 ways to solve this. The simplest is to take the H+ coming from the strong acid as the initial amount of H+ and see how the self ionization of water is affected.

.. .. .. .. .. H2O <=> H+ + .. ... .. . ...OH-
Initial .. .. .. .. .. .. .. 3.26*10^-8
Dissociate .x
Produce .. .. .. .. .. .. x .. .. . .. .. .. .. .. x
At Equil. .. .. .. .. x+3.26*10^-8 .. .. .. x

Kw= [H+][OH-]= (x+3.26*10^-8)x =10^-14 =>
x^2 + 3.26*10^-8x - 10^-14 =0
x=8.50*10^-8

so pH= -log((8.50+3.26)*10^-8)=6.929 =6.93

Answer 2

I'd love for chemgeek to explain how (s)he got a basic pH from a solution of a strong acid.

In this case, you have to add the [H+] coming from the acid to that from the autodissociation of water, so the total [H+] concentration is 1.0*10^-7 + 3.26*10^-8 or 1.33*10^-7. The pH is 6.88.

Answer 3

the contribution of H+ from water would be relatively under a million.0*10^-7 evaluate the Kw equilibrium with H+ from HCl blanketed. (instruments disregarded for readability) H2O (l) <==> H+ (aq) + OH- (aq) I a million*10–8 M 0 C +X +X ---------------------------------------... E (a million*10–8 + X) X understanding that Kw = [H+][OH-] = a million*10^-14 remedy a million*10^-14 = (a million*10–8 + X) X X = - a million.05*10^-7 and 9.5*10^-8 use the 9.5*10^-8 fee (constructive) plug into => (a million*10–8 + X) [H+] = a million*10–8 + 9.5*10^-8 = a million.05*10^-7 pH = -log(H+) = -log(a million.05*10^-7) = 6.97829 = 6.ninety seven or 6.ninety 8 spectacular me if i'm incorrect

Answer 4

this is a strong acid so the [H+] = 3.26 x 10-8

pH = - log [H+] = - log (3.26 x 10-8) = 7.5