[Twoacross]

Just before I post my attempt, the question came with a hint:

The Correct Answer is not 7.35. So im guessing since it is a strong acid, it would completely ionize and finding pH is easy but not in this case it seems.

So my attempt was using the ka value of HNO3, which i found was 28, probably wrong but i continued. From there i calculated kb from:
         
           kb = kw / ka, where kw is 1.00 x 10^-14 so this gave me 3.57 x 10 ^ -16

So from here i set an ice table, when i look at the compound of HNO3, i said that the H ion was inert, so my focus was on the nitrate ion.

So my ice table looks like this:

     NO3^-      +     H2O <=>      OH^- +     HNO3
I 4.5 x 10^-8                             0              0
C      -x                                    +x             +x
E 4.5 x 10^-8-x                          x               x

so here i calculated for x when putting the question in terms of x and i ended up with a pH of 2.602. My teacher told me that this was wrong so i need help on where i could have gone wrong or the correct method to this question

All Help is appreciated

Thank you!

[Borek]

Hint: WHY is 7.35 wrong?

What is theoretical pH of a pure water?

[Twoacross]

I know that 7.35 is wrong since that would assume HNO3 is basic. My problem is that how would I solve this numerically? Such as calculating the pH?

Thanks!

[enahs]

So, to help you with Boreks hint. Neutral water has a pH of 7.0. What is the concentration of [H+] and [OH-] of neutral water?

[Twoacross]

After some calculations, would the concentration of H^+ and OH^- be the same in neutral water?

I did some thinking, if i calculated the concentration of the H ion in the water, which was 1.0 x 10^-7 and added the concentration of the H ion from nitric acid, that would give me the total H concentration, then take the -log of that value and it ended up giving me a pH of 6.84

Could someone confirm or correct me if needed?

Thanks!

[AWK]

This is crude approximation in the right direction.
Exact solution uses K_w in the form
(c+x) x = K_w where c is the concentation of strong acid, x - concnetrations of H⁺ and OH^- from dissociation of water.
solve for x and calculate the total H⁺ concentration c+x.
Show you result and compare with pH=4.84 6.84

[Borek]

I did some thinking, if i calculated the concentration of the H ion in the water, which was 1.0 x 10^-7 and added the concentration of the H ion from nitric acid, that would give me the total H concentration, then take the -log of that value and it ended up giving me a pH of 6.84

You are finally on the right track, even if not there yet. See

http://www.chembuddy.com/?left=pH-calculation&right=pH-strong-acid-base