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