Question

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. ?

Answer 1

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

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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.