Calculate the pH of 0.15M hydrazoic acid, HN3

Question

. Ka = 1.8 x 10-5
.

Answer 1

...............HN3 ==> H^+ + N^3-
I..............0.15..........0.........0
C...............-x............x..........x
E..............0.15-x......x...........x
Plug the E line (equilibrium line) into the Ka expression and solve for x = (H^+).
Then convert H^+ to pH by pH = - log (H^+). Post your work if you get stuck.