Answer:- pH is 2.14.
Solution:- Nitrous acid, [tex]HNO_2[/tex] is a weak acid so first of all we solve for [tex]H_3O^+[/tex] and then figure out the pH.
the equation is written as:
[tex]HNO_2(aq)+H_2O(l)\leftrightharpoons H_3O^+(aq)+NO_2^-(aq)[/tex]
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:
[tex]HNO_2=0.120-x[/tex]
[tex]H_3O^+[/tex] = [tex]x[/tex]
[tex]NO_2^-[/tex] = [tex]x[/tex]
Ka for nitrous acid is [tex]4.5*10^-^4[/tex] and the equilibrium expression for this would be written as:
[tex]Ka=\frac{[H_3O^+][NO_2^-]}{HNO_2}[/tex]
Let's plug in the values in it.
[tex]4.5*10^-^4=\frac{(x)^2}{0.120-x}[/tex]
To make the calculations easy we could ignore [tex]x[/tex] for the bottom and the expression becomes:
[tex]4.5*10^-^4=\frac{(x)^2}{0.120}[/tex]
On cross multiply:
[tex]x^2=4.5*10^-^4*0.120[/tex]
On taking square root to both sides:
[tex]x=7.3*10^-^3[/tex]
So, [tex][H_3O^+]=7.3*10^-^3M[/tex]
Now we could calculate the pH using the pH formula:
[tex]pH=-log[H_3O^+][/tex]
[tex]pH=-log(7.3*10^-^3)[/tex]
pH = 2.14
So, the pH of 0.120M nitrous acid is 2.14.
