Question #62737

As you know, formic acid, #"HCOOH"#, is a weak acid, which means that it does not dissociate completely in aqueous solution.

Simply put, not all molecules of formic acid will actually donate their acidic proton to the water to increase the concentration of hydronium ions, #"H"_3"O"^(+)#, and subsequently decrease the pH of the solution.

You can get an idea about how low the pH of the resulting solution will be by taking a look at the magnitude of the acid dissociation constant, #K_a#, and at the initial concentration of the acid.

Because #K_a# is so small and the solution has such a low a concentration, you can expect the pH of the solution to be much closer to #7# than to #1#. Remember, you'll still be looking at an acidic solution, so you definitely need to have #"pH" < 7#!

Your tool of choice here will be an ICE table

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

#color(purple)("I")" " " "0.0025" " " " " " " " " " " " " " " " " " " " "0" " " " " " " " " " " " " "0#
#color(purple)("C")" " " "(-x)" " " " " " " " " " " " " " " " " "(+x)" " " " " " " "(+x)#
#color(purple)("E")" "0.0025-x" " " " " " " " " " " " " " " " " "x" " " " " " " " " " " "x#

By definition, the acid dissociation constant will be equal to

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

In your case, you will have

#K_a = (x * x)/(0.0025 - x) = 3.5 * 10^(-8)#

Now, because #K_a# is so very small, you can use the following approximation

#0.0025 - x ~~ 0.0025#

This will get you

#K_a = x^2/0.0025 = 3.5 * 10^(-8)#

Solve this equation for #x# to get

#x = sqrt( 0.0025 * 3.5 * 10^(-8)) = 9.35 * 10^(-6)#

Since #x# represents the equilibrium concentration of the hydronium ions, you have

#["H"_3"O"^(+)] = x = 9.35 * 10^(-8)"M"#

and thus

#"pH" = - log( ["H"_3"O"^(+)])#

#"pH" = - log( 9.35 * 10^(-6)) = color(green)(5.03)#