Formic acid (HCOOH) is secreted by ants. Calculate [H3O+] for a 1.18E-2 M aqueous solution of formic acid (Ka = 1.80E-4).

Question

aaronq · Answer

write an equilibrium expression for the dissociation of formic acid. Plug in your values and solve for $[H_3O^+]$.

anonymous · Answer

Which one would I use as the products/reactants?

anonymous · Answer

Would it be HCOOH <==> H30+HCOO?

anonymous · Answer

Giving me, [H30][HCOO]/[HCOOH]?

aaronq · Answer

yes, thats right.

anonymous · Answer

so 1.8E-4=[H30][HCOO]/1.18E-2

1.8E-4*1.18E-2=[H3O][HCOO]

2.124E-6=[H3O][HCOO]

then what?

aaronq · Answer

$[H_3O^+]=[HCOO^-]=x$;  because each dissociation yields 1 molecule of each. 

so $2.124*10^{-6}=x^2$

solve for x

anonymous · Answer

$$\sqrt{2.124e-6}=x$$

x=.001457??

aaronq · Answer

yep, thats it.