Question

Determine the pH at the equivalence (stoichiometric) point in the titration of 20 mL of 0.2 M hydrazine(aq) with 0.14 M HCl(aq). The Kb of hydrazine is 1.7 x 10-6. I am trying to use the equation: ph=7+1/2(pka+logC), but I am unsure how to find the values and plug them in correctly. The answer should be 4.66. Please show all steps clearly. Thanks!

Answer 1

Determine the pH at the equivalence (stoichiometric) point in the titration of 20 mL of 0.2 M hydrazine(aq) with 0.14 M HCl(aq). The K_b of hydrazine is 1.7 x 10^-6. I am trying to use the equation: ph=7+1/2(pka+logC), but I am unsure how to find the values and plug them in correctly. The answer should be 4.66. Please show all steps clearly. Thanks!

The reaction between Hydrazine and HCl is N2H4+HCl ------>N2H5Cl, 1 mole of N2H4 requires 1 mole of HCl at equivalenece point

moles of Hydrazine in 20ml of 0.2M= Molarity* Volume in Liters= 0.2*20/1000 = 0.004, moles of HCl hence= 0.004, molarity of HCl =0.14, Volume of HCl = moles of HCl/Molarity= 0.004/0.14=0.029L= 0.029*1000ml= 29ml

Volume of mixture after reaction = 20+29=49ml= 49/1000L =0.049L

moles of N2H4Cl formed = 0.004, its concentration =0.004/0.049=0.082M

N2H4Cl+H2O-------> N2H4+H3O+

Ka= [N2H4] [H3O+]/[NH4Cl], Ka= 10^-14/Kb= 10^-14/ 1.7*10^-6= 5.88*10^-9

let x= drop in concentration of N2H4Cl to reach equilibrium

at Equilibrium, [N2H4] = [H3O+]= x and N2H4Cl =0.082-x

hence x²/(0.082-x)= 5.88*10^-9, when solved using excel, x= 2.2*10^-5

pH= -log [H3O+]= 4.66