Calculate the pH of the following solutions of HNO3. You must show all work in d

Follow this tutorial, hope it helps.

Solution version 1

HNO3 is a strong acid
HNO3 = H+ + NO3-

we can NOT neglet the autoionization of water :
H2O <=> H+ + OH- for which Kw = 1.0 x 10^-14
because [H+][OH-]= 1.0 x 10^-14
[H+]= 1.0 x 10^-7 M

total concentration of H+ = 1.0 x 10^-7 + 4.8 x 10^-8=1.5 x 10^.7 M
pH = 6.8

Solution version 2

If you use the standard approach for solving pH problems you get:
pH = -log [H+]
pH = -log 4.8*10^-8
pH =7.32
This is technical nonsense because a solution of HNO3 in water is acidic ( naturally) and so it MUST have pH <7.00. What is the problem? And the solution?

This is the method for solving such a problem:
With such dilute solutions of HNO3 you have to take into account the [H+] coming from the ionisation of the water plus the [H+] from the acid
Let the [H+] from the water br X. The [OH-] from the water will also be X
We know that Kw = [H+]*[OH-] = 10^-14 for the water
The total [H+] in the solution = {(4.8*10^-8)+X}
Substituting into the Kw equation:
{(4.8*10^-8)+X} *X = 10^-14
Multiply out and rearrange:
X² + (4.8*10^-8)X - 10^-14 = 0
Solve the quadraric for X: X = 9.998*10^-5
X = [H+] from the water = 9.998*10^-5
[H+] from the acid = 4.8*10^-8
Total [H+] = (9.998+10^-8) + (4.8*10^-8) = 1.4798 * 10^-7
Now:
pH = -log [H+]
pH = -log (1.4798*10^-7)
pH = 6.83

thanks