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Chem 2

posted by on .

Checking to see if I did this right?

Titration of HN03 titrated with NaOH
ending result is 1.38.

Titration A, endpoint pH = 8 is pKa or pKb= 1.38x106=5.861 and

Titration of NH03 titrated with KOH ending results 1.54.

Titration B, endpoint pH = 7 is KaorKb=
Ka=1.54x105

  • Chem 2 - ,

    Beats me. I don't know what you're describing.

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