A 23.8 mL sample of 0.263 M ammonia, NH3, is titrated with 0.217 M perchloric acid....
A 23.8 mL sample of 0.263 M
ammonia, NH3, is
titrated with 0.217 M perchloric
acid.
At the equivalence point, the pH is ____?
Homework Answers
Number of Millimoles = Molarity × Volume (ml)
Millimoles of ammonia used = Millimoles of ammonium perchlorate formed
Total volume after equivalence point = 28.85+23.8 =52.65ml
See the picture below for pH of solution at equivalence point .
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A 23.8 mL sample of 0.263 M
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