the resulting solution will have a pH of what ?

Question

Hydrazine (NH₂NH₂) has a K_b = 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

the correct answer is <7

can you explain why

Answer 1

Hey there!

Let's first find how many moles of each solution we have so that we can do a neutralization table (which always has to be done in moles).

0.1 L * 0.5 M NH₂NH₂ = 0.05 mol NH₂NH₂

0.1 L * 0.5 M HCl = 0.05 mol HCl

Let's write a neutralization equation and plug in our mole values

NH₂NH₂ + HCl ----> NH₂NH₃⁺ + Cl^-

Initial 0.05 mol 0.05 mol 0 0

Change -0.05 mol -0.05 mol +.05 mol + .05 mol

Endpoint 0 mol 0 mol .05 mol .05 mol

The Cl- is a spectator ion, so we can disregard it. The NH₂NH₃⁺, however, is now a weak acid that can dissociate according to its Ka Value. We have to solve for that Ka value:

Ka * Kb = Kw

Ka = Kw / Kb = (1 x 10^-14)/(3 x 10^-6) = 3.33 x 10^-9

Now we write a dissociation equation (ICE tables are done in Mol/L so we need to take the moles of NH₂NH₃⁺ and divide by total volume (0.2 L):

M of NH₂NH₃⁺ = .05 mol/0.2 L = .25 M

NH₂NH₃⁺ <-------> NH₂NH₂ + H⁺

Initial 0.25 M 0 0

Change -x +x +x

Equilibrium 0.25 - x x x

We can disregard the -x on the left side because we have less than 5% dissociation:

Ka = x²/[HA]

3.33 x 10^-9 = x²/0.25

x² = 8.325 x 10^-10

x = [H⁺] = 2.88 x 10^-5

Now, pH = -log (2.88 x 10^-5) = 4.35