A 30.00 mL sample of 0.1500 M hydroazoic acid (HN3; Ka = 1.9 x 10-5) is titrated with 0.1000 M KOH. Calculate the pH after the following volumes have been added:

a. 0.00 mL
b. 11.25 mL
c. 22.50 mL
d. 33.75 mL
e. 45.00 mL
f. 56.25 mL

There is a tremendous amount of typing here. Here is a summary of what you do. Respond by letting me know how many of these you know to do and where your big problem is.

..........HN3 + KOH ==> KN3 + H2O
The first thing you need to do is to determine the mL needed to reach the end point. You want to know if you are before, at, or after the equivalence point.

mols HN3 = M x L = ?
mols KOH = mols HN3
M KOH = mols KOH/L KOH. You know M and mols KOH, solve for L KOH and convert to mL. This is the volume KOH needed to reach the equivalence point. I calculate 45 mL.

a. is the beginning of the titration. This is done as any weak acid or weak base. You've probably done hundreds of these.
............HN3 --> H^+ + 3N^-
I.........0.1500M....0......0
C...........-x......x......x
E.......0.1500-x.....x......x

Substitute the E line into Ka expression and solve for H^+. Convert to pH.

b,c,d. All are before the equivalence point so they are buffered solutions. Use the Henderson-Hasselbalch equation to solve each of these.

d. The equivalence point. The pH is determined by the hydrolysis of the salt. The salt is NaN3 and it's the nitride ion (N^3-) that hydrolyzes. The salt concn is mols/L = 0.0045/0.075 = approx 0.06M

.......N3^- + HOH ==> HN3 + OH^-
I....0.06.............0......0
C......-x.............x......x
E...0.06-x............x......x

Kb for N^3- = (Kw/Ka for HN3) = (HN3)(OH^-)/(N3^-)
Substitute the E line into Kb for N3^- and solve for x = OH^- and convert to H^+ then to pH.

f part is after the equivalence point and is simply the excess OH^- at that point. Convert to H^+ then to pH.

To calculate the pH at each point, we need to determine the concentration of H+, which is related to the concentration of the acid and the volume of the base added.

Here are the steps to calculate the pH at each point:

1. Calculate the initial moles of hydroazoic acid (HN3):
Moles of HN3 = concentration of HN3 (M) × volume of HN3 (L)

For a 30.00 mL sample of 0.1500 M HN3:
Moles of HN3 = 0.1500 M × 0.0300 L

2. Determine the moles of KOH added:
Moles of KOH = concentration of KOH (M) × volume of KOH (L)

3. Calculate the moles of hydroazoic acid (HN3) remaining after each volume of KOH is added:
Moles of HN3 remaining = initial moles of HN3 - moles of KOH

4. Calculate the volume of the remaining solution after each volume of KOH is added:
Volume of remaining solution = initial volume of HN3 - volume of KOH added

5. Calculate the new concentration of hydroazoic acid (HN3):
Concentration of HN3 = moles of HN3 remaining / volume of remaining solution

6. Calculate the hydronium ion (H+) concentration:
For the reaction HN3 + H2O ⇌ H3O+ + N3-, the H+ concentration is equal to the HN3 concentration.

7. Calculate the pH using the formula:
pH = -log[H+]

Now let's plug in the values and calculate the pH at each point:

a. 0.00 mL:
- There is no KOH added, so the concentration and volume of the solution remain the same.
- Moles of HN3 remaining = moles of initial HN3 = 0.1500 M × 0.0300 L
- Concentration of HN3 = moles of HN3 remaining / volume of remaining solution = 0.1500 M
- pH = -log(0.1500)

b. 11.25 mL:
- Moles of KOH = 0.1000 M × 0.01125 L
- Moles of HN3 remaining = initial moles of HN3 - moles of KOH
- Volume of remaining solution = initial volume of HN3 - volume of KOH added
- Concentration of HN3 = moles of HN3 remaining / volume of remaining solution
- Calculate H+ concentration and then the pH using the formula.

c. Repeat the above steps for 22.50 mL, 33.75 mL, 45.00 mL, and 56.25 mL.