Question #27fca

Before doing any calculation, try to predict what you expect the concentrations of #"H"_2"A"# and #"A"^(2-)# to be relative to each other.

Notice that he acid dissociation constant for the first ionization, #K_(a1#, if about three orders of magnitude larger than the acid dissociation constant for the second ionization, #K_(a2)#.

Moreover, the value of #K_(a2)# is very, very small to begin with. This means that you can expect to have a very, very small concentration of #"A"^(2-)# when equilibrium sets in.

In addition to this, you an expect almost all of the hydronium ions, #"H"_3"O"^(+)#, present in solution at equilibrium to come from the first ionization of the acid.

With this in mind, set up an ICE table for the first ionization of the acid

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

#color(purple)("I")" " " "0.0800" " " " " " " " " " " " " " " "0" " " " " " " " " " " "0#
#color(purple)("C")" " " "(-x)" " " " " " " " " " " " " "(+x)" " " " " " " "(+x)#
#color(purple)("E")" "0.0800-x" " " " " " " " " " " " " "x" " " " " " " " " " " "x#

By definition, the acid dissociation constant will be equal to

#K_(a1) = (["H"_3"O"^(+)] * ["HA"^(-)])/(["H"_2"A"]) = 3.6 * 10^(-6)#

This is equivalent to

#3.6 * 10^(-6) = (x * x)/(0.0800 - x)#

Since #K_(a1)# is so small compared with #0.0800#, you can use the approximation

#0.0800 - x ~~ 0.0800#

This will get you

#3.6 * 10^(-6) = x^2/0.0800#

#x = sqrt(0.0800 * 3.6 * 10^(-6)) implies x = 5.37 * 10^(-4)#

Since #x# represents the concentration of hydronium ions produced by the first ionization of the acid and the concentration of #"HA"^(-)#, you will have

#["H"_2"A"] = "0.0800 M" - 5.37 * 10^(-4)"M" = "0.079466 M"#

#["H"_3"O"^(+)] = 5.37 * 10^(-4)"M"#

#["HA"^(-)] = 5.37 * 10^(-4)"M"#

Now, you could skip the second ionization altogether, since you have such a small starting concentration for #"HA"^(-)#, but let's go through with it just to test if our initial estimations were correct.

Use these concentrations to set up a second ICE table, this time of the second ionization of the acid

#" " "HA"_text((aq])^(-) " "+ "H"_2"O"_text((l]) " "rightleftharpoons" " "H"_3"O"_text((aq])^(+) " "+" " "A"_text((aq])^(2-)#

#color(purple)("I")" "5.37 * 10^(-4)" " " " " " " " " " " "5.37 * 10^(-4)" " " " " " " " "0#
#color(purple)("C")" "(-x)" " " " " " " " " " " " " " " "(+x)" " " " " " " " "(+x)#
#color(purple)("E")" "5.37 * 10^(-4)-x" " " " " " " "5.37 * 10^(-4)+x" " " " " "x#

This time, the acid dissociation constant will be

#K_(a2) = (["H"_3"O"^(+)] * ["A"^(2-)])/(["HA"^(-)]) = 6.8 * 10^(-9)#

This will be equivalent to

#6.8 * 10^(-9) = ((5.37 * 10^(-4) + x) * x)/(5.37 * 10^(-4) - x)#

Once again, because #K_(a2)# has such a small value compared with #5.37 * 10^(-4)#, you can say that

#5.37 * 10^(-4) + x ~~ 5.37 * 10^(-4)#

#5.37 * 10^(-4) - x ~~ 5.37 * 10^(-4)#

This will give you

#6.8 * 10^(-9) = (color(red)(cancel(color(black)(5.37 * 10^(-4)))) * x)/color(red)(cancel(color(black)(5.37 * 10^(-4)))) = x#

Therefore, the equilibrium concentration of hydronium ions will be

#["H"_3"O"^(+)] = 5.37 * 10^(-4) + 6.98 * 10^(-9) ~~ 5.37 * 10^(-4)"M"#

As predicted, the first ionization determined the equilibrium concentration of hydronium ions.

The pH of the solution will be

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

#"pH" = - log(5.37 * 10^(-4)) = color(green)(3.27)#

The concentrations of #"H"_2"A"# and #"A"^(2-)# will be

#["H"_2"A"] = color(green)("0.0795 M")#

#["A"^(2-)] = color(green)(6.80 * 10^(-9)"M")#

I'll leave these rounded to three sig figs.