# Question 27fca

Feb 9, 2016

$\text{pH} = 3.27$

#### Explanation:

Before doing any calculation, try to predict what you expect the concentrations of $\text{H"_2"A}$ and ${\text{A}}^{2 -}$ 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, ${K}_{a 2}$.

Moreover, the value of ${K}_{a 2}$ is very, very small to begin with. This means that you can expect to have a very, very small concentration of ${\text{A}}^{2 -}$ when equilibrium sets in.

In addition to this, you an expect almost all of the hydronium ions, ${\text{H"_3"O}}^{+}$, 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

${\text{ " "H"_2"A"_text((aq]) + "H"_2"O"_text((l]) " "rightleftharpoons" " "H"_3"O"_text((aq])^(+) " "+" " "HA}}_{\textrm{\left(a q\right]}}^{-}$

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

${K}_{a 1} = \left(\left[\text{H"_3"O"^(+)] * ["HA"^(-)])/(["H"_2"A}\right]\right) = 3.6 \cdot {10}^{- 6}$

This is equivalent to

$3.6 \cdot {10}^{- 6} = \frac{x \cdot x}{0.0800 - x}$

Since ${K}_{a 1}$ is so small compared with $0.0800$, you can use the approximation

$0.0800 - x \approx 0.0800$

This will get you

$3.6 \cdot {10}^{- 6} = {x}^{2} / 0.0800$

$x = \sqrt{0.0800 \cdot 3.6 \cdot {10}^{- 6}} \implies x = 5.37 \cdot {10}^{- 4}$

Since $x$ represents the concentration of hydronium ions produced by the first ionization of the acid and the concentration of ${\text{HA}}^{-}$, 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 ${\text{HA}}^{-}$, 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

${\text{ " "HA"_text((aq])^(-) " "+ "H"_2"O"_text((l]) " "rightleftharpoons" " "H"_3"O"_text((aq])^(+) " "+" " "A}}_{\textrm{\left(a q\right]}}^{2 -}$

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

${K}_{a 2} = \left(\left[{\text{H"_3"O"^(+)] * ["A"^(2-)])/(["HA}}^{-}\right]\right) = 6.8 \cdot {10}^{- 9}$

This will be equivalent to

$6.8 \cdot {10}^{- 9} = \frac{\left(5.37 \cdot {10}^{- 4} + x\right) \cdot x}{5.37 \cdot {10}^{- 4} - x}$

Once again, because ${K}_{a 2}$ has such a small value compared with $5.37 \cdot {10}^{- 4}$, you can say that

$5.37 \cdot {10}^{- 4} + x \approx 5.37 \cdot {10}^{- 4}$

$5.37 \cdot {10}^{- 4} - x \approx 5.37 \cdot {10}^{- 4}$

This will give you

$6.8 \cdot {10}^{- 9} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{5.37 \cdot {10}^{- 4}}}} \cdot x}{\textcolor{red}{\cancel{\textcolor{b l a c k}{5.37 \cdot {10}^{- 4}}}}} = x$

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"^(+)]))#

$\text{pH} = - \log \left(5.37 \cdot {10}^{- 4}\right) = \textcolor{g r e e n}{3.27}$

The concentrations of $\text{H"_2"A}$ and ${\text{A}}^{2 -}$ will be

$\left[\text{H"_2"A"] = color(green)("0.0795 M}\right)$

$\left[\text{A"^(2-)] = color(green)(6.80 * 10^(-9)"M}\right)$

I'll leave these rounded to three sig figs.