$CH_2H_3O_2(aq) + H_2O(l) rightleftharpoons H_3O^+ + C_2H_3O_2^-(aq)$. $K_c = 1.8 X 10^-5$ at 25 $"^oC$. If a solution initially contains 0.210 M $HC_2H_3O_2$, what is the equilibrium concentration of $H_3O^+$ at 25 $"^oC$?

Md Ariful Islam

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Md Ariful Islam

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Set up an ICE table based on concentrations in mol/l:

$sf(" "CH_2H_3O_(2)+H_2OrightleftharpoonsC_2H_3O_2^(-)+H_3O^(+))$

$sf(color(red)(I)" "0.210" "0" "0)$

$sf(color(red)(C)" "-x" "+x" "+x)$

$sf(color(red)(E)" "(0.210-x)" "x" "x)$

Applying the equilibrium law:

$sf(K_c=(x^2)/((0.210-x))=1.8xx10^(-5))$

Because the dissociation is so small we can make the approximation:

$sf((0.210-x)rArr0.210)$

$:.$$sf(x^(2)=0.210xx1.8xx10^(-5)=0.378xx10^(-5))$

$:.$$sf(x=sqrt(0.378xx10^(-5))=1.9xx10^(-3)color(white)(x)"mol/l")$

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