1
Question
Equilibrium constant for the following reaction is 1×10−9:C5H5N(aq.)+H2O(l)⇌C5H5NH+(aq.)+OH−(aq.)Determine the mole of pyridirium chloride (C5H5N.HCl) that should be added to 500 mL solution of 0.4 M pyridine (C5H5N) to obtain a buffer solution of pH=5.
Open in App
Solution
C5H5N(aq)+H2O⇌C5H5NH(aq)+OH−(aq)
[OH−]=10−pOH=10−1(14−5)=10−9
Ka=[C5H5NH][OH−][C5H5N]=10−9
∴10−9=[C5H5NH]×10−90.4⇒[C5H5NH]=0.4∴molesofC5H5NHCl=0.4×500×10−3=0.2
C5H5N(aq)+H2O⇌C5H5NH(aq)+OH−(aq)
[OH−]=10−pOH=10−1(14−5)=10−9
Ka=[C5H5NH][OH−][C5H5N]=10−9
∴10−9=[C5H5NH]×10−90.4
⇒[C5H5NH]=0.4
∴molesofC5H5NHCl=0.4×500×10−3=0.2
Suggest Corrections
0
Similar questions
View More
Join BYJU'S Learning Program
Explore more
Join BYJU'S Learning Program
