Confused about this problem

Question

Hydrazine (N2H4) is one of the combustible components in rocket fuel and also behaves as a monoprotic base with a 𝐾𝑏 = 1.3 × 10^-6 A researcher tries to characterize rocket fuel by identifying the concentration of hydrazine in the fuel sample. To perform this titration, 50.0 mL of rocket fuel is titrated with 1.50 M HCl.

a. The equivalence point of the titration, as determined by an indicator color

change, was found to occur after adding 31.27 mL of 1.50 M HCl. Determine

the initial concentration of hydrazine in the 50.0 mL sample of rocket fuel.

b. Determine the pH of the analyte solution

1. After adding 20.00 mL of titrant

2. At the equivalence point

3. After adding 40.00 mL of titrant

Answer 1

Not sure where the confusion arises. Did you attempt to solve this problem?

N₂H₄ + HCl ==> N₂H₅⁺ + Cl^-

(a). moles HCl added = 31.27 mls x 1 L / 1000 mls x 1. 50 mol / L = 0.046905 mols HCl

moles N₂H₄ present = 0.046905 mol HCl x 1 mol HCl / mol N₂H₄ = 0.046905 mols N₂H₄

Initial [N₂H₄] = 0046905 mols / 50.0 mls x 1000 mls / L = 0.9381 M

(b1). 20.00 ml HCl x 1 L / 1000 ml x 1.5 mol / L = 0.030 mols HCl added

HCl + N₂H₄ ==> N₂H₅⁺ + Cl^-

0.046905 mol N₂H₄ - 0.030 mols = 0.016905 mols N₂H₄ + formation of 0.030 mols N₂H₅⁺

Final volume = 20 ml + 31.27 ml = 51.27 ml

This forms a buffer: Henderson Hasselbalch equation to find pH:

[N₂H_4] = 0.016905 mol / 51.27 ml x 1 L / 1000 ml = 0.3297 M

[N₂H₅⁺] = 0.030 mol / 51.27 ml x 1 L / 1000 ml = 0.5851 M

pH = pKa + log [conj.base] / [acid]

Ka = 1x10^-14/Kb = 1x10^-14/1.3x10^-6 = 7.69x10^-9

pKa = -log Ka = 8.11

pH = 8.11 + log (0.3297 / 0.5851 = 8.11 - 0.25

pH = 7.86

(b2). At equivalence point in all 0.046905 mols N₂H₄ have been converted to 0.046905 mols N₂H₅⁺

Use Ka of 0.046905 mols N₂H₅⁺ (7.69x10^-9) and the hydrolysis of N₂H₅⁺ to find pH

(b3). After 40 mls, there will be excess HCl (40.00 ml - 31.27 ml = 8.73 ml) and final volume = 71.27 mls.

Find [HCl] and take negative log to get pH.