Question

Question #27fca

Answer 1

`"pH" = 3.27`

Explanation:

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.