Question

 

Calculate the pH for each of the following cases in the titration of 25.0 mL of 0.140 M pyridine, C5H5N(aq) with 0.140 M HBr(aq):

(a) before addition of any HBr

(b) after addition of 12.5 mL of HBr

(c) after addition of 20.0 mL of HBr

(d) after addition of 25.0 mL of HBr

(e) after addition of 37.0 mL of HBr

Answer 1

Answer – The following are the pH with respective addition of HBr-

a) before addition of any HBr

We are given, 25.0 mL of 0.140 M C₅H₅N

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

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

I     0.140                        0                0

C    -x                            +x             +x

E    0.140-x                     + x            +x

Kb for C₅H₅N = 1.7 *10^-9

Kb = [C₅H₆N⁺] [OH^-] / [C₅H₅N]

1.7 *10^-9 = (x) * (x) / (0.140-x)

The x in the 0.190-x can be neglected due to Kb value is too small

1.7 *10^-9 *0.140 = x²

x² = 2.38*10^-10

   x = 1.54*10^-5 M

x = [OH^-] = 1.54*10^-5 M

pOH = -log [OH^-]

         = -log 1.54*10^-5 M

     = 4.81

pH = 14 –pOH

      = 14- 4.81

      = 9.19

We also get , [C₅H₆N⁺] = 1.54*10^-5 M

b) after addition of 12.5 mL of HBr

We need to calculate moles of each

C₅H₅N = 0.140 * 0.025 L = 0.0035 mole

moles of C₅H₆N⁺ = 1.54*10^-5 M * 0.025 L = 3.86*10^-7 moles

When we added 12.5 mL of 0.140 M HBr, then and decrease the number of moles of base and added moles of conjugate base

Moles of HBr = 0.140 M *0.0125 L = 0.00175 moles

New moles of C₅H₅N = 0.0035 - 0.00175 = 0.00175 moles

Moles of C₅H₆N⁺ = 3.86*10^-7 moles + 0.00175 = 0.00175 moles

New molarity-

Total volume = 25 +12.5 = 37.5 mL

[C₅H₅N] = 0.00175 moles / 0.0375 L

              = 0.0467 M

[C₅H₆N⁺ ] = 0.00175 mole / 0.0375 L = 0.0467 M

When base and it conjugate acid have same molarity then there is

pOH = pKb

pKb = - log Kb

pKb = - log 1.7*10^-9

        = 8.77

pH = 14-pOH

      = 14 – 8.77

       = 5.23

c) after addition of 20.0 mL of HBr

so moles of HBr

Moles of HBr = 0.140 M *0.020 L = 0.00280 moles

New moles of C₅H₅N = 0.0035 - 0.00280 = 0.0007 moles

Moles of C₅H₆N⁺ = 3.86*10^-7 moles + 0.00280 = 0.00280 moles

New molarity-

Total volume = 25 +20.0 = 45 mL

[C₅H₆N] = 0.0007 moles / 0.045 L

              = 0.0155 M

[C₅H₆N⁺ ] = 0.00280 mole / 0.045 L = 0.06222 M

Now we need to use Henderson Hasselbalch equation –

pOH = pKb + log [C₅H₆N⁺] / [C₅H₆N]

         = 8.77 + log (0.0622/0.0155)

         = 8.77 + 0.602

         = 9.37

pH = 14- pOH

       = 14 – 9.37

        = 4.63

d) after addition of 25.0 mL of HBr

so we need to calculate moles of HBr

Moles of HBr = 0.140 M *0.025 L = 0.0035 moles

New moles of C₅H₅N = 0.0035 - 0.0035 = 0.0 moles

Moles of C₅H₆N⁺ = 3.86*10^-7 moles + 0.0035 = 0.0035 moles

New molarity-

Total volume = 25 +25 = 50 mL

There are no moles of base, so only moles of conjugate acid

[C₅H₆N⁺ ] = 0.0035mole / 0.050 L = 0.07 M

We need to put ICE chart

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

I     0.07                            0              0

C    -x                             +x             +x

E 0.07-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 / 0.07 –x

x² = 5.88*10^-6 * 0.07

x = 0.000642 M

x = [H₃O⁺] = 0.000642 M

pH = - log [H₃O⁺]

   = - log 0.000642 M

      = 3.19

e) after addition of 37.0 mL of HBr

Moles of HBr = 0.140 M *0.037 L = 0.00518 moles

Moles of C₅H₆N⁺ = 3.86*10^-7 moles + 0.00518 = 0.00518 moles

New molarity-

Total volume = 25 +37 = 62 mL

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

[C₅H₆N⁺ ] = 0.00518 mole / 0.062 L = 0.0835 M

We need to put ICE chart

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

I     0.0835                         0              0

C    -x                              +x             +x

E 0.0835-x                      + x            +x

Ka = [C₅H₅N] [H₃O⁺] / [C₅H₆N]

5.88*10^-6 = x *x / 0.0835 –x

­x² = 5.88*10^-6 * 0.0835

x = 0.000701 M

x = [H₃O⁺] = 0.000701 M

pH = - log [H₃O⁺]

     = - log 0.000701 M

    = 3.15