Hydrazine, N2H4(aq), is used in the preparation of polymers, pharmaceuticals and rocket fuel. Hydrazine has alkaline properties similar to ammonia and will act as a weak base. Given that a 100 mL sample of 0.10 mol/L hydrazine solution has a pH of 10.55 at 25.0 oC, write the formula for the conjugate acid and calculate the Kaof the conjugate acid.

I understand that we have to write the equation for the conjugate acid but I don't understand how to find the values then. I'm hoping someone can help me using the ICE (Initial, Change, Equilibrium) table.

Thank you so much!

I just needed help starting it off and this helped a lot!!

We cant consider the change in the concentration of the N2H4 negligible so do we have to use the quadratic formula to derive Kb?

Given the value of the question I doubt this is the correct way to go about solving this problem..

To write the equation for the conjugate acid of hydrazine, we need to know its chemical reaction with water. Hydrazine can react with water to form its conjugate acid and hydroxide ions:

N2H4(aq) + H2O(l) ⇌ NH3OH+(aq) + OH-(aq)

From this reaction, we can see that the conjugate acid of hydrazine is NH3OH+.

Now, let's use the ICE table to calculate the Ka of the conjugate acid. The initial concentration of NH3OH+ is zero because it is formed from hydrazine in the reaction. We can represent this as:

[NH3OH+]initial = 0

The change in concentration of NH3OH+ is x because some of it will dissociate to produce NH3OH+ ions and OH- ions. The equilibrium concentration of NH3OH+ is also x.

[NH3OH+]change = x
[NH3OH+]equilibrium = x

The initial concentration of OH- ions is given by the equation:

[OH-]initial = 10^(-pOH)

Since the pH of the solution is 10.55, the pOH can be calculated as:

pOH = 14 - pH
pOH = 14 - 10.55
pOH = 3.45

[OH-]initial = 10^(-3.45)

Next, we can write the equation for the ionization of NH3OH+:

NH3OH+(aq) + H2O(l) ⇌ NH4+(aq) + OH-(aq)

At equilibrium, the concentration of NH3OH+ and OH- ions will be equal. Hence, the concentration of OH- ions is also x.

[OH-]equilibrium = x

The equilibrium concentration of NH4+ ions can be calculated using the equation:

[NH4+]equilibrium = [OH-]initial - [OH-]equilibrium

[NH4+]equilibrium = 10^(-3.45) - x

Using the equation for Ka, which is the equilibrium constant for the ionization reaction:

Ka = ([NH4+][OH-])/[NH3OH+]

We can substitute the equilibrium concentrations into this equation:

Ka = ([10^(-3.45) - x] * x) / x

Since the value of x is expected to be very small compared to 10^(-3.45), we can simplify the equation to:

Ka = 10^(-3.45)

Finally, we can calculate the value of Ka using a calculator:

Ka = 10^(-3.45)
Ka ≈ 3.16 x 10^(-4)

Therefore, the formula for the conjugate acid of hydrazine is NH3OH+ and the Ka of the conjugate acid is approximately 3.16 x 10^(-4).

Technically it has two ionizations but the second one is so weak we can ignore it.

pH = 10.55 so pOH = 3.45. Convert that to OH^. That is about 3E-4 but that's just an estimate.
..................N2H4 + H2O ==> N2H5^+ + OH^-
I.................0.1...........................0.................0
C.................-x............................x.................x
E...............0.1-x..........................x..................x
and you know OH^- and N2H5^+ from the above.

Write the Kb expression for N2H4 and substitute the E line and solve for Kb N2H4.
Calculate Ka for the conjugate acid from KaKb = Kw = 1E-14. Post your work if you get stuck.