Question

Determine the pH of a .19 m solution of pyridinium nitrate (C5H5NHNO3) at 25oC. [Pyridinium nitrate dissociates in water to give pyridinium ions (C5H5NH+), the conjugate acid of pyridine (Kb= 1.7x10^-9), and nitrate ions (NO3-).]

Answer 1

@ParthKohli

Answer 2

@Photon336

Answer 3

@Cuanchi

Answer 4

this looks like a weak base in solution will produce OH-

calculate

\[[OH-]= \sqrt{K _{b} \times C _{b} } \]

Answer 5

\[\sqrt{1.7*10^{-9} *19}\]

Answer 6

So something like that?

Answer 7

be carefull!! 0.19 M not 19 M

Answer 8

Ooops sorry. I meant .19

Answer 9

\[\sqrt{1.7*10^{-9} * .19}\]

Answer 10

yea then you calculate the log and subtract from 14 to get the pH

Answer 11

=1.797x10^-5

Answer 12

for the log does it need ot be -log or just log?

Answer 13

- log [OH-]= pOH
pH+pOH= 14

Answer 14

pH =-9.25

Answer 15

Does that seem right?

Answer 16

no, the pH has to be positive!!!!

Answer 17

So 9.25

Answer 18

sounds fine to me!

Answer 19

Thanks