#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#?

1 Answer
Oct 15, 2016

Answer:

#sf([H_3O^(+)]=1.9xx10^(-3)color(white)(x)"mol/l")#

Explanation:

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")#