I need help i do not know how to do this problem at all.
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:
You didn't give any volumes.
First, determine the volume for the equivalence point.
HN3 + KOH ==> KN3 + H2O
mL HN3 x M HN3/mL KOH = mL KOH needed to reach the equivalence point. Then divide the titration into segments as follows: a at the beginning, c at the equivalence point, b all points between a and c, then d is everything past the eq pt.
a. at the beginning you have pure HN3. Treat this as any weak acid.
.........HN3 ==> H^+ + N3^-
I.......0.15.....0......0
C........-x......x......x
E......0.15-x....x......x
Then substitute the E line into Ka expression and solve for x = (H^+), then convert to pH.
b. For all KOH volumes BEFORE the eq. pt. use the Henderson-Hasselbalch equation.
c. For the eq. pt the pH is determined by the hydrolysis of the salt KN3.You will need to determine the concn of the salt which I will call C.
........N3^- + HOH ==> HN3 + OH^-
I.......C...............0....
C......-x...............x......x
E......C-x..............x......x
Kb for N3^- = (Kw/Ka for HN3) = (x)(x)/(C-x) and solve for x = (OH^-), then convert to H^+ and pH.
d. For all points past the eq pt pH is determined by the excess KOH added to the system.
Post your work if you get stuck and show the volumes. Explain fully what you don't understand.