Respuesta :
Answer:
The pH of the solution is 2.56.
Explanation:
Given :
Concentration (c) = 0.40 M
Acid dissociation constant = [tex]K_a=1.9\times 10^{-5}[/tex]
The equilibrium reaction for dissociation of [tex]HN_3[/tex] (weak acid) is,
[tex]HN_3+H_2O\rightleftharpoons N_3^{-}+H_3O^+[/tex]
initially conc. c 0 0
At eqm. [tex]c(1-\alpha)[/tex] [tex]c\alpha[/tex] [tex]c\alpha[/tex]
Dissociation constant is given as:
[tex]K_a=\frac{(c\alpha)(c\alpha)}{c(1-\alpha)}[/tex]
[tex]1.9\times 10^{-5}=\frac{(0.40\alpha)(0.40\alpha)}{0.40(1-\alpha)}[/tex]
By solving the terms, we get value of [tex](\alpha)}[/tex]
[tex]\alpha=0.00686832[/tex]
No we have to calculate the concentration of hydronium ion or hydrogen ion.
[tex][H^+]=c\alpha=0.4\times 0.00686832=0.002747 M[/tex]
Now we have to calculate the pH.
[tex]pH=-\log [H^+][/tex]
[tex]pH=-\log (0.002747)[/tex]
[tex]pH=2.56[/tex]
Therefore, the pH of the solution is, 2.56.
The pH of the aqueous solution of hydrazoic with acid dissociation constant [tex]1.9\;\times\;10^{-5}[/tex] is 2.56.
What is acid dissociation constant?
The acid dissociation constant for the reaction is given as [tex]K_a[/tex]. The acid dissociation constant for hydrazoic acid is [tex]1.9\;\times\;10^{-5}[/tex].
The reaction for the dissociation of acid is given as:
[tex]\rm HN_3\;+\;H_2O\rightarrow\;N_3^-+H_3O^+[/tex]
The initial concentration of acid is 0.40 M. The concentration of reaction at equilibrium is given in the ICE table attached.
The acid dissociation constant for the reaction is given as:
[tex]K_a=\rm \dfrac{[N_3^-][H_3O^+]}{[HN_3]}[/tex]
Substituting the values of concentration:
[tex]1.9\;\times\;10^{-5}=\dfrac{[x][x]}{[0.4-x]} \\x=2.747\;\times\;10^-^3[/tex]
The concentration of hydronium ion or hydrogen ion in the reaction is [tex]x=2.747\;\times\;10^-^3\;\rm M[/tex].
The pH of the reaction can be given as:
[tex]\rm pH=-log\;[H^+]\\pH=-log\;[2.747\;\times\;10^{-3}]\\pH=2.56[/tex]
The pH of the hydrazoic acid solution will be 2.56.
Learn more about acid dissociation constant, here:
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