Answer : The [tex]H_3O^+[/tex] ion concentration is, [tex]1.12\times 10^{-3}M[/tex] and the pH of a buffer is, 2.95
Explanation : Given,
[tex]K_a=7.1\times 10^{-4}[/tex]
Concentration of [tex]HNO_2[/tex] (weak acid)= 0.26 M
Concentration of [tex]KNO_2[/tex] (conjugate base or salt)= 0.89 M
First we have to calculate the value of [tex]pK_a[/tex].
The expression used for the calculation of [tex]pK_a[/tex] is,
[tex]pK_a=-\log (K_a)[/tex]
Now put the value of [tex]K_a[/tex] in this expression, we get:
[tex]pK_a=-\log (7.1\times 10^{-4})[/tex]
[tex]pK_a=4-\log (7.1)[/tex]
[tex]pK_a=3.15[/tex]
Now we have to calculate the pH of the solution.
Using Henderson Hesselbach equation :
[tex]pH=pK_a+\log \frac{[Salt]}{[Acid]}[/tex]
[tex]pH=pK_a+\log \frac{[KNO_2]}{[HNO_2]}[/tex]
Now put all the given values in this expression, we get:
[tex]pH=3.15+\log (\frac{0.89}{0.26})[/tex]
[tex]pH=2.95[/tex]
The pH of a buffer is, 2.95
Now we have to calculate the [tex]H_3O^+[/tex] ion concentration.
[tex]pH=-\log [H_3O^+][/tex]
[tex]2.95=-\log [H_3O^+][/tex]
[tex][H_3O^+]=1.12\times 10^{-3}M[/tex]
The [tex]H_3O^+[/tex] ion concentration is, [tex]1.12\times 10^{-3}M[/tex]