Question

Calculate the pH in an aqueous 0.120 M nitrous acid solution.

Answer 1

Answer:-  pH is 2.14.

Solution:- Nitrous acid, $HNO_2$ is a weak acid so first of all we solve for $H_3O^+$ and then figure out the pH.

the equation is written as:

$HNO_2(aq)+H_2O(l)\leftrightharpoons H_3O^+(aq)+NO_2^-(aq)$

Initial concentration for the acid is given as 0.120 M. Let's say the change in concentration is x. Then the equilibrium concentrations would be as:

$HNO_2=0.120-x$

$H_3O^+$ = $x$

$NO_2^-$ = $x$

Ka for nitrous acid is $4.5*10^-^4$ and the equilibrium expression for this would be written as:

$Ka=\frac{[H_3O^+][NO_2^-]}{HNO_2}$

Let's plug in the values in it.

$4.5*10^-^4=\frac{(x)^2}{0.120-x}$

To make the calculations easy we could ignore $x$ for the bottom and the expression becomes:

$4.5*10^-^4=\frac{(x)^2}{0.120}$

On cross multiply:

$x^2=4.5*10^-^4*0.120$

On taking square root to both sides:

$x=7.3*10^-^3$

So, $[H_3O^+]=7.3*10^-^3M$

Now we could calculate the pH using the pH formula:

$pH=-log[H_3O^+]$

$pH=-log(7.3*10^-^3)$

pH = 2.14

So, the pH of 0.120M nitrous acid is 2.14.

Answer 2

pH can be calculated from pKa of aniline hydrochloride