Question

What is the pH of the solution obtained when 25.0 mL of 0.065M benzylamine, C7H7NH2, is titrated to the equivalence point with 0.050M HCl? Kb=4.7*10^-10. Part 2: What is the pH at the midpoint?

Answer 1

Given :

Volume of benzylamine = 0.025 L

Molarity of benzylamine = 0.065 M

Molarity of HCl = 0.050 M

Kb = 4.7 E-10

Calculation of ka

Ka of C7H7NH3+ = 1.0 E-14 / kb = 1.0 E-14 / 4.7 E-10

= 2.13 E-5

pH at equivalence point :

Calculation of moles of benzylamine

= Molarity x volume in L

= 0.025 L x 0.065 M

= 0.001625 mol

At equivalence point moles of acid = moles of base

Number of moles of HCl = 0.001625 mol

Lets calculate volume of HCl

Volume of HCl = mol / Molarity

=0.001625 mol / 0.050 M

= 0.0325 L

Total volume = 0.025 L + 0.0325 L =0.0575 L

reaction and ICE

C7H7NH2(aq) + HCl (aq) --- > C7H7NH3Cl (aq))

I           0.001625            0.001625              0

C          - 0.001625         -   0.001625              +0.001625

E           0                             0                            0.001625

[C7H7NH3Cl ] = mol /Total volume

= 0.001625 mol / 0.0575 L

=0.02826 M

Since C7H7NH3+ is weak acid and it will dissociates again in the water.

Lets show the reaction

            C7H7NH3+ (aq) + H2O(l) ------ > C7H7NH2 (aq) + H3O+ (aq)

I           0.02826                                               0                      0

C          -x                                                         +x                        +x

E          (0.02826 –x)                                           x                           x

Ka = [C7H7NH2] [H3O+]/[ C7H7NH+]

2.13 E-5 = x² / (0.02826 –x)

Lets assume 5 % approximation.

2.13 E-5 = x² / 0.02826

2.13E-5 x 0.02826 = x²

6.01 E-7 = x²

x = 0.000775

= 7.75 E-4 M

pH = -log [H3O+] = - log ( 7.75 E-4)

=3.11

pH = 3.11

         

pH at mid point :

pH = pka + log ( [base]/[acid])

Total volume is same so number of moles of acid equal to moles of base

So log 1 = 0

At this point pH = pka

pH = -log (2.13 E-5 ) = 4.67