Question

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

Answer 1

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

Answer 2

Which one would I use as the products/reactants?

Answer 3

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

Answer 4

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

Answer 5

yes, thats right.

Answer 6

so 1.8E-4=[H30][HCOO]/1.18E-2

1.8E-4*1.18E-2=[H3O][HCOO]

2.124E-6=[H3O][HCOO]

then what?

Answer 7

\([H_3O^+]=[HCOO^-]=x\); because each dissociation yields 1 molecule of each.

so \(2.124*10^{-6}=x^2\)

solve for x

Answer 8

\[\sqrt{2.124e-6}=x\]

x=.001457??

Answer 9

yep, thats it.