# Chemistry

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

1. 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.

posted by DrBob222

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