pH of 50nM HCl with 2 decimal digits accuracy?

Question

Please justify your answer/provide the calculations.

As a graduate chemist I want to see how many people will get it wrong, so don't worry about cheating with homework ... I am just curious how many people really understand pH.

Answer 1

Kw = 10^-14=[H+][OH-]
so (10^-7+5x10^-8-x)(10^-7-x)=10^14

x=2.19224^-8 so [H+]=....
pH=6.89253 (correct using Debye-Huckel)=6.89271

Answer 2

Oops... Let me try again:

Strong acid in water.
There are three possible scenarios.
1.) The concentration of strong acid is much, much greater than 1x10^-7M, and the autodissociation of water can be ignored in calculating the total [H+].
2.) The concentration of strong acid is much, much lower than 1x10^-7, and the autodissociation of water is more important and the strong acid can be ignored.
3.) The concentratio of strong acid is close to 1x10^-7, and both the strong acid and the autodissociation of water must be considered.

50 nM is 0.5x10^-7, so this problem falls into the third scenario.

HCl is 100% dissociated, and immediately forms 0.5x10^-7 M H+ in solution.

The autodissociation of water will then establish equilibrium around this presence of acid.

H2O.......<- ->........H+.........+.....OH-
I. ... 0.5x10^-7 0
C ... +x +x
E ... 0.5x10^-7+x x

Now, solving the equilibrium at 25 degrees C:
1.00x10^-14 = (0.5x10^-7 + x)(x)
Using quadratic formula and choosing the positive root gives:
x = 7.8078x10^-8
At equilibrium, [H+] = 0.5x10^-7 + x = 1.28078x10^-7
pH = -log[H+] = 6.89
pH = 6.89

Answer 3

the pH of 50nM HCl would be 6.892 or just 6.89

Answer 4

6.82 = -log[1.5X10^-7 M H+]