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.
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.
4 answers
Thank you so much!
I just needed help starting it off and this helped a lot!!
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..
1 answer