Question

In the titration of 25 mL of 0.100 M N2H2 (Kb=8.9x10-7) with .100M HCl, determine the pH at: Please show all steps/ Help please asap

a) the begining of titration

b) after 15 mL of acid has been added

c) after 25 mL of acid has been added

d) after 30 mL of acid has been added

Answer 1

Answer – We are given, 25.0 mL of 0.100 M N₂H₂ , Kb = 8.9*10^-7

a) the begining of titration

we need to put ICE chart

      N₂H₂+ H₂O ------> N₂H₃⁺ + OH^-

I     0.100                    0            0

C    -x                        +x          +x

E 0.100-x                 + x            +x

We know, Kb for N₂H₂= 8.9 *10^-7

Kb = [N₂H₃⁺] [OH^-] / [N₂H₂]

8.9 *10^-7 = (x) * (x) / (0.100-x)

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

8.9 *10^-7 *0.100 = x²

x² = 8.9*10^-8

   x = 2.98*10^-4 M

x = [OH^-] =2.98*10^-4 M

pOH = -log [OH^-]

         = -log (2.98*10^-4 M)

          = 3.52

pH = 14 –pOH

      = 14- 3.52

      = 10.47

[N₂H₃⁺] = x = 2.98*10^-4 M

b) After addition of 15 mL of HCl

moles of N₂H₂= 0.100 M * 0.025 L = 0.0025 mole

moles of N₂H₃⁺ = 2.98*10^-4 M * 0.025 L = 7.46*10^-6 moles

When we added 15 mL of 0.100 M HCl

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

Moles of HCl = 0.100 M *0.015 L = 0.0015 moles

New moles of N₂H₂= 0.0025 - 0.0015 = 0.0010 moles

Moles of N₂H₃⁺ = 7.46*10^-6 moles + 0.0015 = 0.0015 moles

New molarity calculation, so total volume = 25 +15 = 40 mL

[N₂H₂] = 0.0010 moles / 0.040 L

              = 0.025 M

[N₂H₃⁺ ] = 0.0015 mole / 0.040 L = 0.0375 M

Now we need to use Henderson Hasselbalch equation –

pOH = pKb + log [N₂H₃⁺] / [N₂H₂]

pKb = - log Kb

        = - log 8.9*10^-7

         = 6.05

         = 6.05 + log (0.0375/0.025)

         = 6.05 + 0.176

         = 6.23

pH = 14- pOH

       = 14 – 6.23

        = 7.77

c) After addition of 25.0 mL of HCl

Moles of HCl = 0.100 M *0.025 L = 0.0025 moles

New moles of N₂H₂= 0.0025 - 0.0025 = 0.00 moles

Moles of N₂H₃⁺ = 7.46*10^-6 moles + 0.0025 = 0.0025 moles

New molarity calculation, so total volume = 25 +25 = 50 mL

[N₂H₃⁺ ] = 0.0025 mole / 0.050 L = 0.050 M

     N₂H₃⁺ + H₂O ------> N₂H₂ + H₃O⁺

I     0.050                     0              0

C   -x                        +x             +x

E 0.050-x                  + x            +x

Ka = 1*10^-14 / 8.9*10^-7

        = 1.12*10^-8

Ka = [N₂H₂] [H₃O⁺] / [N₂H₃⁺]

1.12*10^-8= x *x / 0.050 –x

x² = 1.12*10^-8* 0.050

x = 2.37*10^-5 M

[H₃O⁺] = x= 2.37*10^-5 M

pH = - log [H₃O⁺]

     = - log 2.37*10^-5 M

    = 4.62

d) After addition of 30.0 mL of HCl

Moles of HCl = 0.100 M *0.030 L = 0.0030 moles

Moles of N₂H₃⁺ = 7.46*10^-6 moles + 0.0030 = 0.0030 moles

New molarity calculation, so total volume = 25 +30 = 55 mL

[N₂H₃⁺ ] = 0.0030 mole / 0.055 L = 0.0545 M

     N₂H₃⁺ + H₂O ------> N₂H₂ + H₃O⁺

I     0.0545                    0              0

C   -x                        +x             +x

E 0.0545-x                + x            +x

Ka = 1*10^-14 / 8.9*10^-7

        = 1.12*10^-8

Ka = [N₂H₂] [H₃O⁺] / [N₂H₃⁺]

1.12*10^-8= x *x / 0.0545 –x

x² = 1.12*10^-8* 0.0545

x = 2.47*10^-5 M

[H₃O⁺] = x= 2.47*10^-5 M

pH = - log [H₃O⁺]

     = - log 2.47*10^-5 M

    = 4.60