Question

A solution is labeled "0.462 M HN3." What are the following in the solution? Express your answers to three significant figures even when only two are valid. Acid-base table [HN3] = M [H3O1+] = M [N31-] = M pH =

Answer 1

Ka of HN3 = 2.5 x 10^-5

HN3 + H2O   -------------> N3-   + H3O+

0.462                                  0           0

0.462 - x                             x             x

Ka = x^2 / 0.462 - x

2.5 x 10^-5 = x^2 / 0.462 - x

x = 3.386 x 10^-3

[H3O+] = 3.39 x 10^-3 M

[N3-] = 3.39 x 10^-3 M

pH = -log ( 3.39 x 10^-3 M)

pH = 2.47