Question

4. Suppose you titrate 50.00 mL of 1.20 M pyridine, C5H5N, with 0.634 M HCl. (a) How many moles of pyridine are present in the original pyridine solution? (b) What is the initial pH? (c) What volume (in mL) of HCl solution is required to reach the equivalence point? (d) What is the pH at the equivalence point? (e) What is the pH of the solution after the addition of 104 mL of HCl?

Answer 1

Here the given pyridine (C5H5N) is a weak base, which react with the strong acid HCl as-

C5H5N + HCl ---------> C5H5NH+Cl-   (salt)

i.e 1 mole of HCl will neutralize 1 mole of pyridine.

i-

Now given concentration of pyridine solution taken = 1.20 M

Volume of pyridine solution taken = 50.00 mL

That means moles of pyridine present in the solution = concentration * volume

= 1.20 M * 50.00 mL

= 1.20 mol/ 1000 mL * 50.00 mL

= 0.06 mols

ii-

Now initial pH of pyridine is calculated in the follwoing way-

Since C5H5N is a weak base, it dissociate in water as-

C5H5N + H2O ---------> C5H5NH+ (conjugate acid) + OH-

Where the dissociation constant (Kb) for pyridine is  1.7 x 10-9.

Again we know Kb = [C5H5NH+] * [OH-] / [C5H5N]

where all the concentrations are at the time of equilibrium.

Now to find these concentrations lets form the ICE table-

Reaction	C5H5N + H2O --------->	C5H5NH+ (conjugate acid) +	OH-
Initial	1.20 M	0	0
Change	-x	+x	+x
Equlibrium	1.20 - x	x	x

Now putting the values-

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

(1.7 x 10-9) = [x] * [x] / [1.20 - x]

(1.7 x 10-9) * [1.20 - x] = x2  

(2.04‬ x 10-9) - (1.7 x 10-9)x = x2  

x2 + (1.7 x 10-9)x - (2.04‬ x 10-9) = 0

Solving this-

x = 0.000045 M

So we can say, at equilibrium

[OH-] = x =  0.000045 M

So

pOH = -log[OH-] = -log[0.000045] = 4.35

So

pH = 14 - pOH = 14 - 4.35 = 9.65‬

iii-

Now equivalence point is that point where all the moles of base present in the solution is completely neutralized by all the acids added.

Here initial moles of C5H5N taken = 0.06 mols

That means 0.06 mols of HCl will be added to convert it completely into 0.06 mols C5H5NH+Cl-  

So volume of HCl required to reach equivalene point = mols / concentration

= 0.06 mols /0.634 M

= 0.06 mols / 0.634 mol/ 1000 mL

= 94.64 mL

iv-

Now at equivalnce point-

fina volume of solution = 50.00 mL + 94.64 mL = 144.64‬ mL

So Concentration of HCl = initial conc * initial vol / final vol = 0.634 M * 94.64 mL / 144.64‬ mL = 0.41M

Again at equilvalnce point since all the base is neutralized by acid, only salt is presenet. i.e at this point

[C5H5NH+Cl-] = [C5H5NH+] = 0.41M

So the pH at this point is completely due to 0.41M of C5H5NH+

Now C5H5NH+ is an acid which dissociate as-

C5H5NH+ + H2O ---------> H3O+ + C5H5N

where

Ka = Kw / Kb

= (10-14) / (1.7 x 10-9)

= 0.58 x 10-5.

Again Ka = [C5H5N] * [H3O+] / [C5H5NH+]

So the ICE table at this point will be-

Reaction	C5H5NH + H2O --------->	C5H5N +	H3O+
Initial	0.41 M	0	0
Change	-x	+x	+x
Equlibrium	0.41 - x	x	x

Now putting the values-

Ka = [C5H5N] * [H3O+] / [C5H5NH+]

(0.58 x 10-5) = [x] * [x] / [0.41 - x]

(0.58 x 10-5) * [0.41 - x] = x2  

(0.237 x 10-5) - (0.58 x 10-5)x = x2  

x2 + (0.58 x 10-5)x - (0.237‬ x 10-5) = 0

Solving this-

x = 0.00155 M

So we can say, at equilibrium

[H3O+] = x =  0.0015 M

So

pH = -log[H3O+] = -log[0.0015] = 2.8