Answer:
Explanations
Calculate the pH for each of the following cases in thetitration of 25.0 mL of 0.100 M pyridine, C5H5N(aq) with 0.100 MHBr(aq):
a.Before and addition of HBr
b.After addition of 12.5ml HBr
c.After addition of 15ml HBr
d.After addition of 25ml HBr
e.After addition of 33ml HBr
SOLUTION ;;;
Kb of pyridine =1.5*10^-9
a)let the dissociation be x.so,
kb=x^2/(0.1-x)
or 1.5*10^-9=x^2/(0.1-x)
or x=1.225*10^-5
so,
[OH-]=1.225*10^-5
so,
pOH=-log([OH-])
so pH=9.088
b)now this will effectively behave as a buffer
pKb=8.82
so pOH=pKb+log(salt/acid)
=8.82+log((12.5*0.1)/(25*0.1-12.5*0.1))
=8.82
so pH=14-pOH
=5.18
c)again using the same equation as the above,
pOH=pKb+log(salt/acid)
=8.82+log((15*0.1)/(25*0.1-15*0.1))
=9
so pH=14-9
=5
d)now the base is completely neutralised.so,
concentration of the salt formed=0.1/2
=0.05 M
so,
pH=7-0.5pKb-0.5log(C)
=7-0.5*8.82-0.5*log(0.05)
=3.24
e)concentration of H+=(33*0.1-25*0.1)/(33+25)
=0.01379
so pH=-log(0.01379)
=1.86
