Question

Consider the titration of 70 ml of 0.25 M hydrazine, N₂H₄ titrated with 0.35 M HBr. ( K_b= 1.3 x 10^-6).

A. Write the reaction equation for above reaction.

B. Calculate the volume of HBr required to reach the end point.

C. Caculate the pH before addition of any acid.

D. Calculate pH after addition of 20 mL of HBrI.

E. Caculate pH after addition of 25 mL of HBr.

F. Caculate pH at the end point.

G. Calculate pH after addition of 59 mL of HBr.

Answer 1

A. Equation : N2H4 + HBr ---> N2H5+ + Br-

B. moles of N2H4 = 0.25 M x 70 ml = 17.5 mmol

Volume of HBr required to reach end point = 17.5 mmol/0.35 M = 50 ml

C. pH before addition of acid

N2H4 + H2O <==> N2H5+ + OH-

let z amunt og N2H4 has reacted

Kb = [N2H5+][OH-]/[N2H4]

1.3 x 10^-6 = x^2/0.25

x = [OH-] = 5.70 x 10^-4 M

pOH = -log[OH-] = 3.24

pH = 14 - pOH = 10.76

D. pH after 20 ml of HBr is added

[N2H4] remaining = (0.25 M x 70 ml - 0.35 M x 20 ml)/90 ml = 0.12 M

1.3 x 10^-6 = x^2/0.12

x = [OH-] = 3.90 x 10^-4 M

pOH = 3.41

pH = 10.59

E. pH after 25 ml of HBr is added

This is half equivalence point

pH = pKa = 14 - pKb = 8.11

F. pH at end point

[N2H5+] = 0.25 M x 70 ml/120 ml = 0.146 M

N2H5+ + H2O <==> N2H4 + H3O+

Ka = 1 x 10^-14/1.3 x 10^-6 = x^2/0.146

x = [H3O+] = 3.35 x 10^-5 M

pH = 4.47

G. pH after 59 ml of HBr added

[HBr] remaining = (0.35 x 9 ml)/129 ml = 0.024 M

pH = -log[0.024] = 1.61