Answer:
Equilibrium concentration of [tex]H^{+}[/tex] is [tex]1.4\times 10^{-3}M[/tex]
Explanation:
Construct an ICE table to calculate changes in concentration at equilibrium.
[tex]CH_{3}COOH\rightleftharpoons CH_{3}COO^{-}+H^{+}[/tex]
I: 0.1 0 0
C: -x +x +x
E: 0.1-x x x
[tex]\frac{[CH_{3}COO^{-}][H^{+}]}{[CH_{3}COOH]}[/tex]= [tex]K_{a}(CH_{3}COOH)[/tex]
(species inside third bracket represent equilibrium concentrations)
[tex]\frac{x^{2}}{0.1-x}=1.85\times 10^{-5}[/tex]
or, [tex]x^{2}+(1.85\times 10^{-5})x-(1.85\times 10^{-6})=0[/tex]
So, [tex]x=\frac{-(1.85\times 10^{-5})+\sqrt{(1.85\times 10^{-5})^{2}+(4\times 1.85\times 10^{-6})}}{2}[/tex]M
or, [tex]x=1.4\times 10^{-3}M[/tex]
So, equilibrium concentration of [tex]H^{+}[/tex]=[tex][H^{+}][/tex]=x=[tex]1.4\times 10^{-3}M[/tex]
The equilibrium concentration will be [tex]H^{+} =1.4\times10^{-3}[/tex]
ICE table for the calculation of changes in concentration at equilibrium.
[tex]CH_{3} COOH= CH_{3}COO^{-1} +H^{+}[/tex]
I: 0.1 0 0
C: -x +x +x
E: 0.1-x x x
[tex]\dfrac{(CH_{3} COO^{-1} ) (H^{+} )}{(CH_{3} COOH)} } =K_{a} \times (CH_{3} COOH)[/tex]
species inside the third bracket represent equilibrium concentrations
[tex]\dfrac{x^{2} }{0.1-x} = 1.85\times10^{-5}[/tex]
[tex]x^{2} +(1.85\times10^{-5} )\times x-(1.85\times10^{-5} )=0[/tex]
[tex]\dfrac{-(1.85\times10^{-5} )+\sqrt{(1.85\times10^{-5})^{2} +(4\times1.85\times10^{-5} ) } }{2}[/tex]
[tex]x=1.4\times10^{-3}[/tex]
The equilibrium concentration will be [tex]H^{+} =1.4\times10^{-3}[/tex]
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