Question

What is the pH of a 0.15 molar solution of HN3 (hydrazoic acid)?

Given: Ka = 1.8 × 10-5

A.
4.56
B.
11.22
C.
1.36
D.
2.78

Answer 1

D. 2.78

Step-by-step explanation:

`HN_3 (aq)+ H_2 O(l) <>N_3^- (aq)+H_3 O^+ (aq)`

Initial 0.15M 0 0

Change -x +x +x

Equilibrium 0.15M-x +x +x

`Ka =\frac {((x)(x))}{(0.15M-x)}`

`1.8*10^(-5)= \frac {x^2}{(0.15M-x)}`

(-x is neglected) so we get

`1.8*10^(-5)*0.15=x^2\\\\x^2=2.7*10^(-6)`

`x=√(x^2)=1.64*10^(-3) M=H^3 O^(+)`

`pH=-log[H^3 O^+]\\\\pH=-log[1.64*10^(-3)]\\\\pH=2.78`

Option D is the Answer