Question

What is the ph of a 0.045 M solution of the weak Baer pyridine?

Answer 1

Solution :-

concentration of the pyridine = 0.045 M

pH= ?

kb of the pyridine = 1.7*10^-9

When pyridine react with water then it forms the OH- ions

C5H5N    + H2O ------- > C5H5NH^+ + OH^-

0.045 M                                0                  0

-x                                        +x                 +x

0.045-x                                x                   x

Lets write the kb equation

Kb= [C5H5NH^+] [OH^-] /[C5H5N]

1.7*10^-9 = [x][x]/[0.045-x]

since kb is very small therefore we can neglect the x from the denominator

1.7*10^-9 = [x][x]/[0.045]

1.7*10^-9 * 0.045 = x^2

7.65*10^-11 = x^2

taking square root of both sides we get

8.75*10^-6 =x = [OH-]

now using the [OH-] concnetration lets calculate the pOH

pOH = -log [OH-]

pOH= - log [8.75*10^-6]

pOH =5.06

now lets find pH

pH+ pOH = 14

therefore pH= 14 - pOH

                  = 14 - 5.06

                  =8.94

Therefore the pH of the solution = 8.94