Question

Hydrazine (NH2NH2) has a Kb = 3.0 x 10-6. If 100.0 mL of a 0.5000 M aqueous hydrazine solution is mixed with 100.0 mL of 0.5000 M aqueous hydrochloric acid, the resulting solution will have a pH A. = 7 OOO B. <7 O C. >> OD. Cannot be predicted from the information given

Answer 1

Given the mixture hydrazine is a weak base as it's Kb is very small and a strong acid HCl .First we have to calculate the number of moles of both weak base and strong acid .We see that the moles of acid is equals to moles of base hence we conclude that the there will be formation of salt as both acid and base are consume in equal amount. Therefore salt form undergo equilibrium with hydrazine and proton form. So we have to find the concentration of salt as we have 0.05 moles and total volume 200 mL.Then we consider let 'x' moles of salt dissociates at equilibrium to give x moles of proton and x moles of hydrazine. We use the Ka (7.7 *10-9 from literature)value for salt (i.e is a weak acid type) and solve the equation to have x .We assume that x is very very small than the initial concentration of acid and ignore it.Then after calculating x we have concentration of proton and crossponding pH. So the pH calculated is less than 7.Therefore B is the correct option

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Hus --- NH₂Nth + Ha Hading + ceo Number of moles of HN-NM = 100|X0.5000 M = 0.05 moles Number of moles of Hue ULX0.5000 M T1000) = 0.05 moles NI-NH₂ + He HN-By+cio 0.05 molen 00 Smoles 0 0 o o.o Smoles 0.05 molen → Concentration of (ND-ROM] - 0.05 (moles) (100+100) 2 1000 – LN4, Ngut = 8.25M Ny-Nh tut Ny-Nz - Initially 0:25 At equilibrium. 10. 25-n) Ka = [NG-ng] [ut] [NH2-NU] 7.78169 0.25-2 0:25 as 0.438x10-4 - 4.38 x 10-5 22 ♡ _x2 0.25-2 pH = - -log[ut] log (4.38x10-5) pH = 4.36 Scanned with CamScanner