Question

Find the pH and the volume (mL) of 0.487 M HNO3 needed to reach the equivalence point in the titration of 2.65 L of 0.0750 M pyridine (C5H5N)?

Answer 1

The number of moles of pyridine in 2.65 L of 0.0750 M of pyridine can be calculated as follows:

Number of moles of pyridine = Molarity x volume

Number of moles of pyridine = 0.0750M x 2.65L

Number of moles of pyridine = 0.199 mol

The balanced equation of reaction of pyridine with nitric acid can be shown as follows:

C5H5N   +   HNO3   <-------> C5H5NH⁺    +   NO3^-

i.e. C5H5N   +   H⁺   <-------> C5H5NH⁺   

Thus, one mole of pyridine reacts with one mole of HNO3.

Therefore, number of moles of HNO3 = number of moles of pyridine = 0.199 mol

Given solution of HNO3 is 0.487M. Therefore, volume of HNO3 solution containing 0.199 mol of HNO3 can be calculated as follows:

Volume of 0.487M HNO3 = Moles / Molarity

Volume of 0.487M HNO3 = (0.199 mol) / (0.487 mol/L)

Volume of 0.487M HNO3 = 0.409 L

Volume of 0.487M HNO3 = 409 mL

Therefore, volume (mL) of 0.487 M HNO3 needed to reach the equivalence point is 409 mL

pH of 409 mL of 0.487 M HNO3 solution can be calculated as follows:

pH = -log (0.487) = 0.312

Thus, pH of HNO3 solution is 0.312

pH at the equivalence point is due to H⁺ ion obtained by the dissociation of C5H5NH⁺ . It can be calculated as follows:

Concentration of C5H5NH⁺ = Moles of C5H5NH⁺ / Volume of solution

Concentration of C5H5NH⁺ = 0.199 mol / (2.65L + 0.409L)

Concentration of C5H5NH⁺ = 0.199 mol / 3.06L

Concentration of C5H5NH⁺ = 0.0650 M

Let ‘s’ be the amount of C5H5NH⁺ dissociated. The ICE chart for dissociation of C5H5NH⁺ can be given as follows:

C5H5NH⁺    <-------> C5H5N   +   H⁺  

Initial                     0.0659M                               0              0

Change                 -s                                            +s           +s

Equilibrium         0.0659M -s                          s              s

Thus, expression for Ka can be written as follows:

Ka = [C5H5N][ H⁺]/[ C5H5NH⁺]

Value of Ka = 5.90 x 10^-6

5.90 x 10^-6 = s² / (0.0659M –s)

Value of s can be assumed to be very small.

5.90 x 10^-6 = s² / 0.0659M

s² = (5.90 x 10^-6) x 0.0659M

s = 0.624 x 10^-3

Therefore, [H+] = 0.624 x 10^-3

Thus, pH = -log [H+]

pH = -log (0.624 x 10^-3)

pH = 3.20

Thus, pH at equivalence point is 3.20