Question

6.) A 50.00 mL sample of 0.1000 M pyridine is titrated with 0.2000 M hydrochloric acid. Show your calculations for the pH of the solution after the addition of each of the four total volumes of 0.2000 M hydrochloric acid in items a-d below.

a.) 0.00 mL

b.) 20.00 mL

c.) 25.00 mL

d.) 26.00 mL

Answer 1

Answer –

a)pH Before addition of HCl

We are given, 50.0 mL of 0.100 M C₅H₅N

It is the weak base, so we need to put ICE chart

      C₅H₅N + H₂O ------> C₅H₆N⁺ + OH^-

I     0.100                         0              0

C    -x                              +x             +x

E 0.100-x                      + x            +x

Kb for C₅H₅N = 1.7 *10^-9

Kb = [C₅H₆N⁺] [OH^-] / [C₅H₅N]

1.7 *10^-9 = (x) * (x) / (0.100-x)

The x in the 0.100-x can be neglected due to Kb value is too small

1.7 *10^-9 *0.100 = x²

x² = 1.7*10^-10

   x = 1.3*10^-5 M

x = [OH^-] = 1.3*10^-5 M

pOH = -log [OH^-]

         = -log (1.3*10^-5 M)

        = 4.88

pH = 14 –pOH

      = 14- 4.88

      = 9.12

We also get , x = [C₅H₆N⁺] = 1.30*10^-5 M

b) We need to calculate moles of C₅H₅N = 0.100 * 0.05 L = 0.005 mole

moles of C₅H₆N⁺ = 1.30*10^-5 M * 0.050 L = 6.52*10^-7 moles

When we added 20.00 mL of 0.200 M HCl

Then there is added moles of conjugate base and decrease the number of moles of base

Moles of HCl = 0.200 M *0.020 L = 0.0040 moles

New moles of C₅H₅N = 0.005 - 0.004 = 0.001 moles

Moles of C₅H₆N⁺ = 6.52*10^-7 moles + 0.004 = 0.004 moles

New molarity-

Total volume = 50 +20 = 70 mL

[C₅H₅N] = 0.001 moles / 0.070 L

              = 0.0143 M

[C₅H₆N⁺ ] = 0.004 mole / 0.070 L = 0.0572 M

Now we need to use Henderson Hasselbalch equation –

pOH = pKb + log [C₅H₆N⁺] / [C₅H₆N]

         = 8.77 + log (0.0572/0.0143)

         = 8.77 + 0.602

         = 9.37

pH = 14- pOH

       = 14 – 9.37

        = 4.63

c) We already calculate moles of C₅H₅N and C₅H₆N⁺ in the part b

so we need to calculate moles of HCl

Moles of HCl = 0.200 M *0.025 L = 0.0050 moles

New moles of C₅H₅N = 0.005 - 0.0050 = 0.00 moles

Moles of C₅H₆N⁺ = 6.52*10^-7 moles + 0.0050 = 0.0050 moles

New molarity-

Total volume = 50 +25 = 75 mL

There is no moles of base, so only moles of HCl

[C₅H₆N⁺ ] = 0.005 mole / 0.075 L = 0.0666 M

We need to put ICE chart

      C₅H₆N⁺ + H₂O ------> C₅H₅N + H₃O⁺

I     0.0666                         0              0

C    -x                              +x             +x

E 0.0666-x                      + x            +x

Ka = 1*10^-14 / 1.7*10^-9

        = 5.88*10^-6

Ka = [C₅H₅N] [H₃O⁺] / [C₅H₆N]

5.88*10^-6 = x *x / 0.0666 –x

x² = 5.88*10^-6 * 0.0666

x = 0.000626 M

x = [H₃O⁺] = 0.000626 M

pH = - log [H₃O⁺]

     = - log 0.000626 M

    = 3.20

d) We already calculate moles of C₅H₅N and C₅H₆N⁺ in the part b

so we need to calculate moles of HCl

Moles of HCl = 0.200 M *0.026 L = 0.0052 moles

New moles of C₅H₅N = 0.005 - 0.0052 = 0.00 moles

Moles of C₅H₆N⁺ = 6.52*10^-7 moles + 0.0052 = 0.0052 moles

New molarity-

Total volume = 50 +26 = 76 mL

There is no moles of base, so only moles of HCl

[C₅H₆N⁺ ] = 0.0052 mole / 0.076 L = 0.0684 M

We need to put ICE chart

      C₅H₆N⁺ + H₂O ------> C₅H₅N + H₃O⁺

I     0.0684                         0              0

C    -x                              +x             +x

E 0.0684-x                      + x            +x

Ka = 1*10^-14 / 1.7*10^-9

        = 5.88*10^-6

Ka = [C₅H₅N] [H₃O⁺] / [C₅H₆N]

5.88*10^-6 = x *x / 0.0684 –x

x² = 5.88*10^-6 * 0.0684

x = 0.000626 M

x = [H₃O⁺] = 0.000634 M

pH = - log [H₃O⁺]

     = - log 0.000634 M

    = 3.197