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%.
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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