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.