[Urgent] Titration Problem!

lazybone

1. The problem statement, all variables and given/known data
Titrate 60ml of .15M of HNO3 with .45M NaOH.
A) Give pH of Initial Point.
B) Give pH of Equivalent Point.
C) Give pH of Mid Point.
D) Give pH of End point.

2. Relevant equations
nacid=nbase


3. The attempt at a solution
A) I think I did this right.
HNO3-> H + NO3
Strong Acid = 100% dissociated
pH= .8
B)nacid=nbase
(.15M)(.06L)=(.45M)x L
= .02L~20ml
C) I'm stuck here. Strong acid+ strong base = 1/2 neutralization.
???/(.06+.02/2) =
I don't get how I get the top part of that equation. The answer was said to be, .064M H+. pH = 1.2.
D) Strong Acid + Strong Base = H20, pH = 7.

Please help!!! I have a test this tuesday.

Borek

You have not answered B, volume is not pH.

C is just an excess reagent questions. Compare this pH calculation question.

D doesn't make much sense as asked. End point depends on the detection method, if you are not told how it was detected, you can't say at what pH it was detected.