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

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SUMMARY

The discussion focuses on calculating the pH at various points during the titration of 60ml of 0.15M HNO3 with 0.45M NaOH. The initial pH is determined to be 0.8, while the equivalent point requires calculating the volume of NaOH needed, resulting in a pH of 1.2 at the mid-point with 0.064M H+. The end point is noted to be pH 7, but the exact pH can vary based on the detection method used. The calculations emphasize the importance of understanding strong acid and strong base interactions in titration.

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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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