Question

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

Answer 1

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..