How would I get the Ice of the titration of 40.00ml of 0.100M H2NNH2(aq) having a Kb of 3.0x10^-6 by 0.100M HClO4(aq)?

Trying to plot a pH curve. I know the combination of acid base is strong acid weak base

millimoles H2NNH2 = mL x M = 40.00 x 0.100M = 4.000 mmoles.

mmoles HClO4 = mL x 0.1M = ??. You don't list any of the volumes.
Here is how the ICE chart would look.

....H2NNH2 + HClO4 ==> H2NNH3^+ + ClO4^-
begin..4.000...0........0...........0
add...........?mmols.................
change.-?mmoles.-?mmoles.+mmoles.
equil..4-?.......0......+mmoles

Then use the Kb expression for zero mL HClO4. Use the Henderson-Hasselbalch equation for all HClO4 volumes after the beginning and before the equivalence point. Use the hydrolysis of the salt at the equivalence point and use excess HClO4 for all volumes after the equivalence point. The curve should look like one of these (or the reverse of one of these).
http://www.chemguide.co.uk/physical/acidbaseeqia/phcurves.html

To find the pH curve of the titration between 40.00 ml of 0.100 M H2NNH2(aq) and 0.100 M HClO4(aq), you would need to calculate the pH at various points of the titration. The titration involves the reaction between a strong acid (HClO4) and a weak base (H2NNH2). Here's how you can do it:

1. Calculate the initial moles of H2NNH2:
- Moles = concentration x volume
- Moles H2NNH2 = 0.100 M x 0.04000 L

2. Calculate the initial moles of HClO4:
- Moles HClO4 = 0.100 M x volume of HClO4 added (let's say V)

3. Determine the limiting reagent:
- The limiting reagent is the one that is completely consumed first. In this case, it will be either H2NNH2 or HClO4, depending on the volumes used in the titration.

4. Calculate the moles of excess and remaining reactant:
- If H2NNH2 is the limiting reagent, then all of it will react, and you will have excess HClO4.
- If HClO4 is the limiting reagent, then it will all react, and you will have excess H2NNH2.

5. Calculate the moles of the excess reactant:
- Excess moles = Initial moles - moles of limiting reagent

6. Determine the volume of acid/base required for the stoichiometric equivalence point:
- Since H2NNH2 and HClO4 react in a 1:1 ratio, the volume of HClO4 added at the equivalence point will be equal to the volume of H2NNH2 initially taken (40.00 ml).

Now that you have determined the stoichiometric equivalence point, you can plot the pH curve by calculating the pH at different volumes of HClO4 added before, during, and after the equivalence point. To do this:

7. Calculate the moles of HClO4 at any point during the titration:
- Moles of HClO4 = 0.100 M x volume of HClO4 added (V)

8. Determine the moles of HClO4 reacted by subtracting the moles of excess HClO4:
- Moles of HClO4 reacted = Moles of HClO4 initially - Moles of excess HClO4

9. Calculate the moles of H2NNH2 remaining by subtracting the moles of H2NNH2 reacted:
- Moles of H2NNH2 remaining = Moles of H2NNH2 initially - Moles of H2NNH2 reacted

10. Use the Henderson-Hasselbalch equation to calculate the pH:
- pH = pKa + log([A-]/[HA]), where [A-] is the concentration of the conjugate base and [HA] is the concentration of the weak acid.
- In this case, [A-] represents the concentration of the H2NNH2 that has reacted with HClO4 to form its conjugate acid (NH4+).
- [HA] represents the concentration of the H2NNH2 remaining.

By repeating steps 7-10 for various volumes of HClO4 added, you can generate a pH curve that shows the changes in pH as the titration progresses.