Given a solution of nitric acid whose #pH=4#, what are #[H_3O^+]#, and #[NO_3^(-)]#?

1 Answer
Dec 30, 2016

Answer:

#[HNO_3]=0.0001*mol*L^-1#

Explanation:

#pH=-log_10[H_3O^+]#

We are given that #pH=4#

And thus #[H_3O^+]=0.0001*mol*L^-1# (since #log_(10)0.0001=log_(10)10^-4=-(-4)=4#.

In water, nitric acid would undergo complete dissociation, according to the following rxn:

#HNO_3(aq) + H_2O(l)rarrH_3O^+ + NO_3^-#

And thus the formal concentration of nitric acid is equivalent to #[H_3O^+]# and #[NO_3^-]#.