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Adipic acid is a diprotic acid having $K_\mathrm {a1}=3.9\times10−5$ and $K_\mathrm {a2}=3.9\times10−6$. A saturated solution of adipic acid is about $\pu{0.10 M}~\ce{HOOC(CH2)4COOH}$. Calculate the concentration of the $\ce{−OOC(CH2)4COO−}$ in this solution. Express your answer to two significant figures and include the appropriate units

Lets represent adipic acid as $\ce{H2A}$ for the sake of simplicity.

$$\ce{H2A <=> H+ + HA-}\qquad K_{\mathrm a1} = 3.9 \times 10^{-5}$$

$$\begin{array}{lccc} \text{Initial}\quad &0.1 \quad &&0 \quad &&0\\ \text{Change}\quad &0.1-x \quad &&+x \quad &&+x\\ \text{Eq}\quad &0.1-x \quad &&x \quad &&x \end{array}$$

$$3.9 \times 10^{-5} = x\times x/0.1-x$$

$x^2 = 3.9 \times 10^{-5} \times 0.1$ (neglecting $x$ in nominator because $x\ll0.1$)

$x = x_1 = 0.00197\ \mathrm M$

Let's designate $x$ as $x_1$. $\ce{HA-}$ dissociates again as

$$\ce{HA- <=> H+ + A-}\qquad K_{\mathrm a2} = 3.9 \times 10^{-6}$$

$$\begin{array}{lccc} \text{Initial}\quad &0.00197 \quad &&0 \quad &&0\\ \text{Change}\quad &0.00197 -x \quad &&+x \quad &&+x\\ \text{Eq}\quad &0.00197 -x \quad &&x \quad &&x \end{array}$$

$$3.9 \times 10^{-6} = x^2/0.00197$$

$x^2 = 0.00197 \times 3.9 \times 10^{-6}$ as shown above

$x = x_2 = 8.77 \times 10^{-5}$ (designate $x$ as $x_2$)

Concentration of $\ce{HOOC(CH2)4COOH}$ ($\ce{H2A}$) $= 0.1 - x_1 = 0.098\ \mathrm M$

But that was wrong. I then $x_1 - x_2 = 0.00197 - 8.77 \times 10^{-5} = 0.00188\ \mathrm M$ but that answer is wrong.

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I'll follow your lead and represent adipic acid as $\ce{H2A}$ for the sake of simplicity.

Equations we know:

(I) o.100 = $\ce{[H2A] + [HA^{-}] + [A^{2-}]}$

(II) $\dfrac{\ce{[H^+] [HA^{-}]}}{\ce{[H2A]}} = 3.9 * 10^{−5}$

(III) $\dfrac{\ce{[H^+] [A^{-2}]}}{\ce{[HA^{-}]}} = 3.9 * 10^{−6}$

(IV) $\ce{[H^+]} = \ce{[HA^{-}] + 2[A^{2-}]}$

There are four equations in four unknowns so the system does have a solution. I could find any way to simply all the equations into anything which seems reasonable for an exact solution.

The way to solve this problem is to solve for the first ionization and assume that the second ionization has an insignificant contribution. A check can be done with equation (IV) to verify that $\ce{[HA^{-}] \gg 2[A^{2-}]}$.

I am going to show an excessive number of significant figures in the answers so that various methods of calculation can be compared. The acid constants are only given to 2 significant figures so that is the limiting number of significant figures that can justified.

Quick and dirty

Let $\ce{[H^+] = [HA^{-}]}$ and let $\ce{[HA]} = 0.100$

then from (II)

$\ce{[H^{+}]^2 = [HA^{-}]^2 =} \sqrt{(0.100)(3.9E-5)}$

$\ce{[H^{+}] = [HA^{-}] =} 1.97E-3$

thus $[\ce{HA}] \approx 0.1 - 0.00197 = 0.09803$

from equation (III)

(V) $\ce{[A^{-2}]} = \dfrac{\ce{[HA^{-}]}}{\ce{[H^{+}]}}3.9E-6 = 3.9E-6$

So indeed $\ce{[HA^{-}] \gg [A^{-2}]}$

This method can be made somewhat better by noting

                  x*(0.1/0.10197)
   HA    0.10000  0.09807
   HA-   0.00197  0.00193
         -------  -------
         0.10197  0.10000

Cleaner solution

The Quick and dirty is a bit dodgy since $\ce{[H2A]}$ was assigned a value of 0.100 to calculate, then reassigned a value of 0.09803. Since the pK constant is only given to two significant figures this isn't really wrong, but a more consistent approach can be easily calculated.

Let $\ce{[H^+] = [HA^{-}]} = x$ and let $\ce{[HA]} = 0.100 - x$, and also let's use $k_2$. Substituting into equation (III):

$x^2 = (0.100 -x)k_2$

$x^2 + k_2x -0.1k_2$

This is a quadratic in x so roots are:

$x = \dfrac{-k_2 \pm \sqrt{k_2^2 - 4(1)(-0.1k_2)}}{2(1)}$

The negative root doesn't work so the positive one is the only one. Also $k_2^2$ is so small that it can be neglected so;

$x = \dfrac{-k_2 + \sqrt{0.4k_2}}{2} =0.00186$

$\ce{[H^+] = [HA^{-}]} = 0.00186$

$\ce{[HA]} = 0.100 - 0.00186 = 0.09814 $

Again from equation (V)

$\ce{[A^{-2}]} = \dfrac{\ce{[HA^{-}]}}{\ce{[H^{+}]}}3.9E-6 = 3.9E-6$

Exact Solution

Solve equation (II) for $\ce{[HA^{-}]}$ in terms of $\ce{[H2A]}$ and $\ce{[H^+]}$ to get:

$\ce{[HA^{-}]} = \dfrac{k_1\ce{[H2A]}}{\ce{[H^+]}}$

and multiply (II) and (III) to solve for $\ce{[HA^{2-}]}$ in terms of $\ce{[H2A]}$ and $\ce{[H^+]}$ to get:

$\ce{[A^{2-}]} = \dfrac{k_1k_2\ce{[H2A]}}{\ce{[H^+]^2}}$

Both of the above can be substituted into the left hand side of the following three equations to get the values shown on the right.

$\dfrac{\ce{[H2A]}}{\ce{[H2A] + [H2A^{-}] + [A^{2-}]}} = \dfrac{\ce{[H^+]^2}}{\ce{[H^+]^2 + k_1[H^+] + k_1k_2}}$

$\dfrac{\ce{[HA^{-}]}}{\ce{[H2A] + [H2A^{-}] + [A^{2-}]}} = \dfrac{\ce{k1[H^+]}}{\ce{[H^+]^2 + k_1[H^+] + k_1k_2}} $

$\dfrac{\ce{[A^{2-}]}}{\ce{[H2A] + [H2A^{-}] + [A^{2-}]}} = \dfrac{\ce{k_1k_2[H^+]}}{\ce{[H^+]^2 + k_1[H^+] + k_1k_2}} $

Being a glutton for punishment I created a spreadsheet with the calculations. After some manual manipulation I was able to determine the "exact solution" to a totally ridiculous number of decimal places.

$\ce{[\ce{H+}]} \text{ }= 0.00195932639309$

$\ce{[\ce{H2A}]}~~= 0.09804455814272$
$\ce{[\ce{HA-}]} ~= 0.00195155732146$
$\ce{[\ce{A^{2-}}]}~~= 0.00000388453582$
$\text{sum} \quad = 0.10000000000000$

$\ce{2[\ce{A^{2-}}] = [\ce{H+}] - [\ce{HA-}]} = 0.00000776907164$

$\ce{K_{a1}}$ from data = $3.9E-5$
$\ce{K_{a2}}$ from data = $3.9E-5$

$\ce{[H^+]} = \ce{[HA^{-}] + 2[A^{2-}]} = 0.00195932639309$

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