What is the pH of this solution after the following amounts of NaOH have been added?

Question

A sample of 50.00 mL of 0.1000 M hydrazoic acid, HN3, is titrated with 0.1000 M NaOH.

What is the pH of this solution after the following amounts of NaOH have been added:

a.) 0.00 mL

b.) 25.00 mL

c.) 50.00 mL

d.) 50.10 mL

Answer 1

Not asking much, are you?

Ka (HN3) = 2.5 * 10^-5 = [H+][(N3)-]/[HN3]

Stated in problem [HN3] = 0.1000M

Substituting:  2.5 * 10^-6 = [H+][(N3)-]

Assuming [H+] (H2O) = 0 and all [H+] is from HN3 => [H+] = [(N3)-]

2.5 * 10^-6 = [H+]^2  ==> [H+] = 1.6 * 10^-3  ==> pH = 2.8

Now, do the rest yourself.