Calculate the pH of HCl of concn 1 x 10 ^ (-8).?

Question

Its an acid remember!!! so the pH has gotta be below 7... HELP!

I came up with the answer 8 and the teacher said "nah mate".... Hydrochloric acid man!!!! its an acid and it has a pH of below 7!!!!

Answer 1

The concentration is so low that you have to take into account the self dissociation of water.
However you DO NOT just add up the H+ from pure water and that from HCl since theself dissociation of water is an equilibrium.

So you either consider that you have the equilibrium of water which is shifted by the addition of HCl or you find how much water will self dissociate in the presence of the H+ from HCl (2 different ways to get to the same result).

In the first case, for pure water [H+]0= [OH-]0 =10^-7
So

.. .. .. .. H2O <=> H+ + OH-
Initial .. .. . .. .. .. .. 10^-7 .. 10^-7
Add .. .. .. .. .. ... .. .. 10^-8 .. ..
React .. .. .. .. .. .. .. .. x .. .. .. x
At Eq .. .. .. .. .. .. 1.1*10^-7 -x.. 10^-7-x
Kw=[H+][OH-]= (1.1*10^-7-x) * (10^-7 -x) =10^-14
solve for x and then
pH= -log(1.1*10^-7-x)

In the second case you have
.. .. .. .. H2O <=> H+ + OH-
Initial .. .. . .. .. .. .. 10^-8
Dissociate .x
Produce .. .. .. .. .. x .. .. .. x
At Eq .. .. .. .. .. 10^-8 +x .. x

Kw=(10^-8+x)x=10^-14

Solve for x
and then pH=-log(10^-8+x)

E.g. in this case x=9.5*10^-8
pH=-log(10^-8 + 9.5*10^-8) =-log(10.5*10^-8) = 6.98

Answer 2

the contribution of H+ from water will be rather below a million.0*10^-7 evaluate the Kw equilibrium with H+ from HCl coated. (contraptions handed over 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 sparkling up 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 value (helpful) 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 properly acceptable me if i'm incorrect

Answer 3

It's a trick question. Remember that pure water has a small amount of hydronium ion, amounting to 1.0 x 10 ^ (-7) M. You add to that the hydronium contributed by HCl, which is 1 x 10 ^ (-8) M, so the total amount of hydronium ion is now 1.1 x 10 ^ (-7). The negative logarithm of that number is the pH, which comes out to 6.96.
In case you're wondering why most problems don't require you to take into account the hydronium contributed by water, it's because normally chemists deal with HCl in concentrations far exceeding 1.0 x 10 ^ (-7). For instance, if the concentration of HCl were 0.01 M, it would be unnecessary to take into account the hydronium from water, because adding 0.0000001 M to 0.01 M won't make a difference.

Answer 4

even though it is an acid, especially HCl, but it does not mean that the pH cannot be more than 7. so pH = - (log(1E-8))

Answer 5

8. no it is a base