# Question 8298c

Mar 27, 2015

So, you have your diprotic acid, ${H}_{2} A$, and the equilibrium reactions that take place in aqueous solution

${H}_{2} {A}_{\left(a q\right)} r i g h t \le f t h a r p \infty n s {H}_{\left(a q\right)}^{+} + H {A}_{\left(a q\right)}^{-}$, $p {K}_{a 1} = 5.9$,

$H {A}_{\left(a q\right)}^{-} r i g h t \le f t h a r p \infty n s {H}_{\left(a q\right)}^{+} + {A}_{\left(a q\right)}^{2 -}$, $p {K}_{a 2} = 9.4$

This means that your solution will contain three species: ${H}_{2} A$, $H {A}^{-}$, and ${A}^{2 -}$. The idea behind speciation is that the total concentration of a solution made by mixing these three species is constant, regardless of what actual proportions you have.

Each of these three species will have a fraction of the total concentration that depends on their respective concentrations. For example,

$\frac{\left[{H}_{2} A\right]}{\left[{H}_{2} A\right] + \left[H {A}^{-}\right] + \left[{A}^{2 -}\right]} = {\text{fraction}}_{{H}_{2} A}$

Now, I can't show you the actual calculations because that would make for a very, very long answer. If you use the definitions of the acid dissociation constants, ${K}_{a 1}$ and ${K}_{a 2}$, you can write the fraction each species has by using the concentration of ${H}^{+}$, and ${K}_{a 1}$ and ${K}_{a 2}$.

These fractions will look like this

$\frac{{\left[{H}^{+}\right]}^{2}}{{\left[{H}^{+}\right]}^{2 +} + {K}_{a 1} \cdot \left[{H}^{+}\right] + {K}_{a 1} \cdot {K}_{a 2}}$ $\to$ fraction of ${H}_{2} A$

$\frac{{K}_{a 1} \cdot \left[{H}^{+}\right]}{{\left[{H}^{+}\right]}^{2 +} + {K}_{a 1} \cdot \left[{H}^{+}\right] + {K}_{a 1} \cdot {K}_{a 2}}$ $\to$ fraction of $H {A}^{-}$

$\frac{{K}_{a 1} \cdot {K}_{a 2}}{{\left[{H}^{+}\right]}^{2 +} + {K}_{a 1} \cdot \left[{H}^{+}\right] + {K}_{a 1} \cdot {K}_{a 2}}$ $\to$ fraction of ${A}^{2 -}$

This form will help you with the actual speciation diagram because you can calculate these fractions for a given pH without too much trouble.

Start with pH = 0, determine $\left[{H}^{+}\right]$ by using

$\left[{H}^{+}\right] = {10}^{- p H}$

and the values of ${K}_{a 1}$ and ${K}_{a 2}$ from

${K}_{a 1} = {10}^{- p {K}_{a 1}}$ and ${K}_{a 2} = {10}^{- p {K}_{a 2}}$

For the entire pH range your diagram will look like this That speciation diagram belongs to ${H}_{2} C {O}_{3}$. Here's how you'd adapt yours.

• The $\textcolor{g r e e n}{\text{green}}$ curve will be $\left[{H}_{2} A\right]$;
• The $\textcolor{red}{\text{red}}$ curve will be $\left[H {A}^{-}\right]$;
• The $\textcolor{b l u e}{\text{blue}}$ curve will be $\left[{A}^{2 -}\right]$;

Now, your $\textcolor{g r e e n}{\text{green}}$ curve will intersect your $\textcolor{red}{\text{red}}$ curve at $p {K}_{a 1} = 5.9$, and your $\textcolor{red}{\text{red}}$ curve will intersect your $\textcolor{b l u e}{\text{blue}}$ curve at $p {K}_{a 2} = 9.4$.

For the intermediate species, in your case $\left[H {A}^{-}\right]$, the pH of the solution will be

$p {H}_{\text{intermediate}} = \frac{p {K}_{a 1} + p {K}_{a 2}}{2} = \frac{5.9 + 9.4}{2} = 7.65 \cong 7.7$

Mar 27, 2015

Here's a sample calculation of you you can derive the fractions formulas that only use $\left[{H}^{+}\right]$, ${K}_{a 1}$, and ${K}_{a 2}$

According to the definitions of ${K}_{a 1}$ and ${K}_{a 2}$, you'll get

(1): ${K}_{a 1} = \frac{\left[{H}^{+}\right] \cdot \left[H {A}^{-}\right]}{\left[{H}_{2} A\right]}$, and

(2): ${K}_{a 2} = \frac{\left[{H}^{+}\right] \cdot \left[{A}^{2 -}\right]}{\left[H {A}^{-}\right]}$

I'll use this notation for simplicity

$\left[{H}_{2} A\right] = x$, $\left[H {A}^{-}\right] = y$, and $\left[{A}^{2 -}\right] = z$.

The fraction of $\left[{H}_{2} A\right]$ will be

${f}_{{H}_{2} A} = \frac{x}{x + y + z}$ (MAIN)

Use (1) to write a value for $x$

$x = \frac{\left[{H}^{+}\right] \cdot y}{K} _ \left(a 1\right)$ (A)

Use (2) to get a value for $y$

$y = \frac{\left[{H}^{+}\right] \cdot z}{K} _ \left(a 2\right)$ (B)

Plug (B) into (A)

$x = \frac{\left[{H}^{+}\right] \cdot \left[{H}^{+}\right] \cdot z}{{K}_{a 1} \cdot {K}_{a 2}} = \frac{{\left[{H}^{+}\right]}^{2} \cdot z}{{K}_{a 1} \cdot {K}_{a 2}}$

Now plug everything into (MAIN) and you'll get

${f}_{{H}_{2} A} = \frac{\frac{{\left[{H}^{+}\right]}^{2} \cdot z}{{K}_{a 1} \cdot {K}_{a 2}}}{\frac{{\left[{H}^{+}\right]}^{2} \cdot z}{{K}_{a 1} \cdot {K}_{a 2}} + \frac{\left[{H}^{+}\right] \cdot z}{K} _ \left(a 2\right) + z}$

${f}_{{H}_{2} A} = \frac{{\left[{H}^{+}\right]}^{2} \cdot z}{{K}_{a 1} \cdot {K}_{a 2} \cdot \left(\frac{{\left[{H}^{+}\right]}^{2} \cdot z}{{K}_{a 1} \cdot {K}_{a 2}} + \frac{\left[{H}^{+}\right] \cdot z}{K} _ \left(a 2\right) + z\right)}$

f_(H_2A) = ([H^(+)]^(2) * cancel(z))/(([H^(+)]^(2) * cancel(z) + K_(a1) * [H^(+)] * cancel(z) + K_(a1) * K_(a2) * cancel(z))

Finally,

${f}_{{H}_{2} A} = \frac{{\left[{H}^{+}\right]}^{2}}{{\left[{H}^{+}\right]}^{2} + {K}_{a 1} \cdot \left[{H}^{+}\right] \cdot {K}_{a 1} \cdot {K}_{a 2}}$

Mar 27, 2015

The speciation diagram is H₂A ⇌ H⁺ + HA⁻; $\text{p} {K}_{1} = 5.9$
HA⁻ ⇌ H⁺ + A²⁻; $\text{p} {K}_{2} = 9.4$

There are three species: H₂A, HA⁻, and A²⁻.

The material balance is given by

$C = \text{[H₂A] + [HA⁻] + [A²⁻]}$, where $C$ is the total concentration of the acid.

The fraction of each species is given by

${f}_{\text{H₂A" = "[H₂A]}} / C$
${f}_{\text{HA⁻" = "[HA⁻]}} / C$
${f}_{\text{A²⁻" = "[A²⁻]}} / C$

I will omit the detailed calculations because, Stefan Zdre has done them for you.

Let $D = \text{[H⁺]"^2 + K_1"[H⁺]} + {K}_{1} {K}_{2}$

${f}_{\text{H₂A" = "[H⁺]}}^{2} / D$; f_"HA⁻" =( K_1"[H⁺]")/D#; ${f}_{\text{A²⁻}} = \frac{{K}_{1} {K}_{2}}{D}$

Now we can use Excel to plot the fraction of each species as a function of pH.

The data are I actually plotted the data at 0.2 pH intervals to get more points. The plots are shown at the beginning of this answer.

Note how the plots cross over at exactly $\text{p} {K}_{1}$ and $\text{p} {K}_{2}$.

The $\textcolor{p u r p \le}{\text{purple}}$ HA⁻ line doesn't make it all the way back up to 1.00 because the $\text{p} K$ values are so close.

But it does peak at pH 7.6, which is the average of $\text{p} {K}_{1}$ and $\text{p} {K}_{2}$.