pH of 10^-8 M HCl?

Question

This question has been asked before but wasn't answered very well. I'd appreciate working, reasons etc.

Thanks!

Answer 1

I've answered this before.

At such low concentrations of HCl the self-dissociation of water has to be considered.
The self ionization of water is an EQUILIBRIUM, thus you cannot simply add 10^-8 (that comes from HCl) and 10^-7 (from water) to find [H+].
There are two possible ways to do it. Of course you will reach the same result but the first one has the simplest calculations.

A) We want to see how the self-dissociation of water occurs in the presence of 10^-8 M H+

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

Note that [H2O] is never included in such cases as it is considered constant and that's why I am not including anything at the initial and equilibrium values for H2O

Kw=[H+][OH-]= (10^-8+x)x =10^-14 =>
x^2 +10^-8x-10^-14=0
x=9.51*10^-8

Thus pH=-log(x+10^-8)= -log(9.51*10^-8+10^-8) = 6.978 =6.98

B) we can treat it as pure water whose equilibrium is shifted by the addition of H+
So you have 10^-7+ 0.1*10^-7 =1.1*10^-7 initial H+
Now the equilibrium will shift to the left because you add H+

.. .. .. .. .. .. .. H2O <=> H+ + .. .. OH-
Initial .. .. .. .. .. .. .. .1.1*10^-7 ... .. 10^-7
React .. .. .. .. .. .. .. .. .. x .. .. .. .. .. x
Produce .. .. .. x
At Equil .. .. .. .. .. . 1.1*10^-7-x .. .. 10^-7-x

Kw= (1.1*10^-7-x)(10^-7-x) =10^-14=>

x^2-2.1*10^-7x+ 10^-15=0
x1= 2.05*10^-7 rejected since it is more than the initial amounts
x2= 4.8751*10^-9
and
pH=-log(1.1*10^-7 -x) =-log(1.1*10^-7 -4.8751*10^-9) =6.978= 6.98

If you need any clarifications please say so.

Answer 2

The pH of 10^-8 M HCl IS 6.98 . Because the molar concentration of HCl is negligible as compared to dissociation of water that is 10^-7 therefore such a concentration of acid ( <10^-7 ) does not have much impact on pH of water.

Answer 3

The H+ ion concentration is (1 x 10-7) + (1 x 10-8).

So add them together and then take the negative logarithm.

The water itself provides more H+ ions than the HCl.

Answer 4

HCl>>H+ +Cl- so if HCl is 10^-8 M the concentration of H+ is 10^-8 m.
But we have to remember the dissociation of H2O
H2O<> H+ + OH- Kw=10^-14
H+=10^-7
Total concentration of H+ = 10^-8+10^-7=1.1 10^-7
pH=6.96
In this case we can not trascure the dissociation of H2O

Answer 5

going strictly by the pH equation (pH = -log[H+]), you get a pH value of 8

BUT

this is a solution in water and water undergoes autoionization where (at 25 deg C) the [H+] will be = 1 e-7

so the TOTAL [H+] = the sum of the two or 1.1 e-7
and the pH = - log(1.1 e-7) = 6.9586 = 7.0

essentially, the wate is providing more H+ than the acid is