Solve Titration Problem: pH of Initial, Equiv., Mid & End Point

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The discussion focuses on solving a titration problem involving 60ml of 0.15M HNO3 titrated with 0.45M NaOH. The initial pH is calculated as 0.8, while the equivalent point occurs at approximately 20ml of NaOH added, requiring further clarification on the calculation method. The mid-point pH is identified as a half-neutralization scenario, with a suggested pH of 1.2 based on 0.064M H+. The end point is noted to be pH 7, but the accuracy of this value is questioned due to the dependence on detection methods. Overall, the thread emphasizes the need for precise calculations and understanding of titration concepts.
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[Urgent] Titration Problem!

Homework Statement


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.

Homework Equations


nacid=nbase

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.
 
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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.
 
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