Be sure to answer all parts. What is the [H3O+] and the pH of a buffer that consists of 0.26 M HNO2 and 0.89 M KNO2? (K, of HNO2 = 7.1 x 10-4) fH 0+1= x 10 M (Enter your answer in scientific notation.) pH =

Respuesta :

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]

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