Respuesta :
Answer : The value of acid dissociation constant is, [tex]1.1\times 10^{-5}[/tex]
Solution : Given,
Concentration pyridinecarboxylic acid = 0.78 M
pH = 2.53
First we have to calculate the hydrogen ion concentration.
[tex]pH=-\log [H^+][/tex]
[tex]2.53=-\log [H^+][/tex]
[tex][H^+]=2.95\times 106{-3}M[/tex]
Now we have to calculate the acid dissociation constant.
The equilibrium reaction for dissociation of (weak acid) is,
[tex]HC_6H_4NO_2\rightleftharpoons C_6H_4NO_2^-+H^+[/tex]
initially conc. 0.78 0 0
At eqm. (0.78-x) x x
The expression of acid dissociation constant for acid is:
[tex]k_a=\frac{[C_6H_4NO_2^-][H^+]}{[C_6H_4NO_2]}[/tex]
As, [tex][H^+]=[C_6H_4NO_2^-]=x[/tex]
So, [tex]x=2.95\times 106{-3}M[/tex]
Now put all the given values in this formula ,we get:
[tex]k_a=\frac{(x)\times (x)}{(0.78-x)}[/tex]
[tex]k_a=\frac{(2.95\times 106{-3})\times (2.95\times 106{-3})}{(0.78-2.95\times 106{-3})}[/tex]
[tex]K_a=1.1\times 10^{-5}[/tex]
Therefore, the value of acid dissociation constant is, [tex]1.1\times 10^{-5}[/tex]
