hmmm this is not making sense at the moment.
the question:
How many milliliters of 0.246 M HNO3 should be added to 213 mL of 0.006 66 M 2,2-bipyridine to give a pH of 4.19?
the equation:
HNO3- + 22bipy --> Protenated bipy + NO3 2-
u want to find the moles of HNO3...
there are 1.1419 mmoles of bipy...
tried doing an ice table,
H A- HA
I 1.1419 1.1419 0
C -x -x +x
E 1.1419-x 1.1419-x x
and then doing the henderson hasselbach eq but it didn't work out...
pkA of bipy is 4.34.
pH = pka + log ([moles A-] / [moles HA])
4.19=4.34 + log [1.1419-x]/x
1 answer
nevermind, all clear on this matter, i guess i should try harder before posting on here haha
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