For this, problem, we will use an ICE table.
To start, we need the net ionic equation for the reaction of #HCOOH#, formic acid, with #H_2O#, water.
#HCOOH(aq)+H_2O(l)rightleftharpoonsH_3O^"+"(aq)+HCOO^"-"(aq)#
Now, we can create our ICE table, where I stands for initial concentration, C stands for change, and E stands for equilibrium concentration.
#color(white)(mmm)HCOOH(aq)+H_2O(l)rightleftharpoonsH_3O^+(aq)+HCOO^"-"(aq)#
#color(white)(l)# -I-#color(white)(mm)1.0color(white)lMcolor(white)(mmmmm)-color(white)(mmml)0color(white)(l)Mcolor(white)(mmmlmm)0color(white)(l)M#
#color(white)(l)# -C-#color(white)(mm)-xcolor(white)(mmmmlm)-color(white)(mlmm)+xcolor(white)(lmmmmm)+x#
#color(white)(l)# -E-#color(white)(llm)1.0-xcolor(white)(mmmlm)-color(white)(mmmlm)xcolor(white)(mlmmmmlm)x#
#color(white)m#
Now, write the #K_a# expression and plug the values in.
#color(white)m#
#K_a=([H_3O^+][HCOO^-])/([HCOOH])=1.78*10^-4=((x)(x))/((1.0-x))#
Solve for #x#. We will ignore the #x# in the denominator, because #x# will most likely be so small that it will be negligible when adding or subtracting.
#x^2/(1.0cancel(color(red)(-x)))=1.78*10^-4#
#x^2=1.78*10^-4#
#x=sqrt(1.78*10^-4)#
#x=1.3*10^-2#
#[H_3O^+]=1.3*10^-2color(white)(l)M#
We can now check if #x# really was negligible when we ignored it before. This can be done by dividing #x# by the equilibrium concentration of #HCOOH#, multiplying by 100, and checking if the resultant value is less than #5%#.
#(1.3*10^-2)/(1.0-1.3*10^-2)=1.32%<5%#
Because #1.32%# is less than #5%#, we can conclude that our choice to ignore #x# earlier in the problem was valid.
