Answer:
D. 2.78
Explanation:
[tex]HN_3 (aq)+ H_2 O(l) <>N_3^- (aq)+H_3 O^+ (aq)[/tex]
Initial 0.15M 0 0
Change -x +x +x
Equilibrium 0.15M-x +x +x
[tex]Ka =\frac {((x)(x))}{(0.15M-x)}[/tex]
[tex]1.8\times10^{-5}= \frac {x^2}{(0.15M-x)}[/tex]
(-x is neglected) so we get
[tex]1.8\times10^{-5}\times0.15=x^2\\\\x^2=2.7\times10^{-6}[/tex]
[tex]x=\sqrt{x^2}=1.64\times10^{-3} M=H^3 O^{+}[/tex]
[tex]pH=-log[H^3 O^+]\\\\pH=-log[1.64\times10^{-3}]\\\\pH=2.78[/tex]
Option D is the Answer
