Respuesta :
Answer:
Percent ionization of HCOOH is 3.69%
Explanation:
To calculate percent ionization of HCOOH, we have to construct an ICE table to determine changes in concentrations at equilibrium.
[tex]HCOOH\rightleftharpoons HCOO^{-}+H^{+}[/tex]
I: 0.125 0 0
C: -x +x +x
E: 0.125-x x x
species inside third bracket represent equilibrium concentrations
So, [tex]\frac{[HCOO^{-}][H^{+}]}{[HCOOH]}=K_{a}(HCOOCH)[/tex]
or, [tex]\frac{x^{2}}{0.125-x}= 1.77\times 10^{-4}[/tex]
or, [tex]x^{2}+0.000177x-0.0000221 = 0[/tex]
[tex]x=\frac{-0.000177+\sqrt{(0.000177)^{2}+(4\times 0.0000221)}}{2}[/tex]
or, [tex]x=4.61\times 10^{-3}[/tex] M
So, [tex][H^{+}]=4.61\times 10^{-3}M[/tex]
Percent ionization of formic acid = [tex]\frac{[H^{+}]}{initial concentration of HCOOH}\times 100[/tex] = [tex]\frac{0.00461}{0.125}\times 100[/tex] = 3.69%
The percent ionization of the acid had been the concentration of hydrogen ion with respect to the acid concentration. The percent ionization of formic acid is 3.69%.
What is the acid dissociation constant (Ka)?
The acid dissociation constant has been the concentration of the acid dissociated into the constituent ions.
The dissociation of formic acid is given as:
[tex]\rm HCOOH\;\rightleftharpoons\;H^+\;HCOO^-[/tex]
The acid dissociation constant (Ka) for formic acid is given as:
[tex]\rm Ka=\dfrac{[H^+][HCOO^-]}{[HCOOH]}[/tex]
Substituting the concentration of the ions and the acid from the ICE table attached.
[tex]\rm 1.77\;\times\;10^{-4}=\dfrac{[x][x]}{[0.125-x]} \\\\x=4.61\;\times\;10^-3[/tex]
The hydrogen ion concentration in the solution has been 0.00461 M.
Substituting the concentration for the percent ionization of the formic acid:
[tex]\rm Percent\;ionization=\dfrac{[H^+]}{[HCOOH]}\;\times\;100\\\\ Percent\;ionization=\dfrac{0.00461}{0.125}\;\times\;100\\\\ Percent\;ionization=3.69\;\%[/tex]
The percent ionization of formic acid is 3.69%.
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