Formic acid (HCOOH), Ka = 1.8 X 10-4, is the irritant associated with ant stings. Calculate the equilibrium concentrations of all species, HA, H+, and A-1, of an aqueous solution containing 18.0 g of formic acid per liter. ?

I got #x = "0.00839 M"#. I'll let you read below how to get the rest. What is the pH of this solution?

Well, I would first find the molarity of the starting acid.

#(18.0 cancel"g HCOOH")/"1 L HCOOH" xx "1 mol"/(46.0248 cancel"g HCOOH")#

#=# #"0.391 M formic acid"#

The reaction and ICE table become:

#"HCOOH"(aq) + "H"_2"O"(l) rightleftharpoons "HCOO"^(-)(aq) + "H"_3"O"^(+)(aq)#

#"I"" "0.391" "" "" "" "-" "" "" "0" "" "" "" "" "" "0#
#"C"" "-x" "" "" "" "-" "" "" "+x" "" "" "" "+x#
#"E"" "0.391-x" "" "-" "" "" "x" "" "" "" "" "" "x#

And so, the #K_a# expression is:

#K_a = (["HCOO"^(-)]["H"_3"O"^(+)])/(["HCOOH"]) = x^2/(0.391 - x)#

The #K_a# is fairly small, so we make the small #x# approximation, i.e. #0.391 - x ~~ 0.391#, so that

#1.8 xx 10^(-4) ~~ x^2/0.391#

Thus,

#x = sqrt(0.391 cdot 1.8 xx 10^(-4)) = "0.00839 M"#

So then, from the ICE table, we find the equilibrium concentrations to be:

#color(blue)(x -= ["H"_3"O"^(+)]_(eq) = ["HCOO"^(-)]_(eq) = ul"0.00839 M")#

#color(blue)(0.391 - x -= ["HCOOH"]_(eq) = ul"0.383 M")#

So was our approximation good? Is #"0.00839 M"/"0.391 M" < 0.05#? Well, I get about #0.01#, so... yes.