In: Chemistry
You carry out the titration of 35.00 mL of 0.1750 M pyridine using a 0.275 M HCl solution. Pyridine has a pKb = 8.77
What is the pH at:
0 ml HCl added
5.0 ml HCl added
half equivalence point for HCl
half equivalence point for pyridine
17.0 ml HCl added
21.0 ml HCl added
equivalence point
26.0 ml HCl added
Here is my answer,
PYRIDINE + H+ ---->pyridineH + (pyridine H+)
1)0 ml HCl
no. of moles of pyridine= 0.035 lit X 0.1750 = 0.006125 moles
no.of moles of HCl=0
pOH=Pkb+log [pyridineH]/[pyridine]
8.77=[x] X [x]/0.1750-x
solve the quadratic equation we get 0.171
pH=log 0.171=2.23
2)5 ml
no. of moles of pyridine= 0.035 lit X 0.1750 = 0.006125 moles
no.of moles of HCl=0.005litX 0.275=0.001375
no.of moles of pyridineH+ =0.005litX 0.275=0.001375
no.of moles of pyridine left= 0.006125-0.001375=0.00475 moles
pOH= 8.77+log (0.001375)/0.00475= 8.77-0.53=8.24
pH=14-8.24=5.76
3)17 ml HCl
no.of moles of HCl=0.017X0.275=0.004675 moles
no.of moles of pyridineH+=0.004675 moles
no.of moles of pyridine left=0.006125-0.004675=0.001450 moles
pOH=8.77+log 0.004675/0.001450=8.77+log 3.22=9.28
pH=14-9.28=4.72
4)21 ml HCl
similarly,
no.of moles of HCl=0.021X0.275=0.005775
no.of moles of pyridineH+=0.005775
no.of moles of pyridine left=0.006125-0.005775=0.00035
pOH=8.77+log 0.005775/0.00035=8.77+1.217=9.98
pH=14-9.98=4.02
5)26 ml of HCl
no.of moles of HCl=0.026X0.275=0.00715
no.of moles of pyridineH+=0.00715
no.of moles of pyridine excess=0.00715-0.006125=-0.00102
pOH=8.77+log 0.00715/0.00102=8.77+.0.84=9.61
pH=14-9.61=4.39
6) at equivalence point the concentration of pyridine is equal to the concentration of HCl
n1v1=n2v2
(35X06.125)/0.275=22.2
no.of moles of HCl=0.0222X0.275=0.006105
pOH=8.77+log0.006105/0.006125=8.77-0.0014=8.76
pH=14-8.76=5.24
6) at half equivalence point. the pH=pKa
pH=8.77
this is the best i could..