Question

Question #8298c

Answer 1

!! LONG ANSWER !!

So, you have your diprotic acid, `H_2A`, and the equilibrium reactions that take place in aqueous solution

`H_2A_((aq)) rightleftharpoons H_((aq))^(+) + HA_((aq))^(-)`, `pK_(a1) = 5.9`,

`HA_((aq))^(-) rightleftharpoons H_((aq))^(+) + A_((aq))^(2-)`, `pK_(a2) = 9.4`

This means that your solution will contain three species: `H_2A`, `HA^(-)`, 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,

`([H_2A])/([H_2A] + [HA^(-)] + [A^(2-)]) = "fraction"_(H_2A)`

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_(a1)` and `K_(a2)`, you can write the fraction each species has by using the concentration of `H^(+)`, and `K_(a1)` and `K_(a2)`.

These fractions will look like this

`([H^(+)]^(2))/([H^(+)]^(2+) + K_(a1)*[H^(+)] + K_(a1) * K_(a2))` `->` fraction of `H_2A`

`(K_(a1) * [H^(+)])/([H^(+)]^(2+) + K_(a1)*[H^(+)] + K_(a1) * K_(a2))` `->` fraction of `HA^(-)`

`(K_(a1) * K_(a2))/([H^(+)]^(2+) + K_(a1)*[H^(+)] + K_(a1) * K_(a2))` `->` 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 `[H^(+)]` by using

`[H^(+)] = 10^(-pH)`

and the values of `K_(a1)` and `K_(a2)` from

`K_(a1) = 10^(-pK_(a1))` and `K_(a2) = 10^(-pK_(a2))`

For the entire pH range your diagram will look like this

That speciation diagram belongs to `H_2CO_3`. Here's how you'd adapt yours.

• The `color(green)("green")` curve will be `[H_2A]`;
• The `color(red)("red")` curve will be `[HA^(-)]`;
• The `color(blue)("blue")` curve will be `[A^(2-)]`;

Now, your `color(green)("green")` curve will intersect your `color(red)("red")` curve at `pK_(a1) = 5.9`, and your `color(red)("red")` curve will intersect your `color(blue)("blue")` curve at `pK_(a2) = 9.4`.

For the intermediate species, in your case `[HA^(-)]`, the pH of the solution will be

`pH_("intermediate") = (pK_(a1) + pK_(a2))/2 = (5.9 + 9.4)/2 = 7.65 ~= 7.7`

Answer 2

Here's a sample calculation of you you can derive the fractions formulas that only use `[H^(+)]`, `K_(a1)`, and `K_(a2)`

According to the definitions of `K_(a1)` and `K_(a2)`, you'll get

(1): `K_(a1) = ([H^(+)] * [HA^(-)])/([H_2A])`, and

(2): `K_(a2) = ([H^(+)] * [A^(2-)])/([HA^(-)])`

I'll use this notation for simplicity

`[H_2A] = x`, `[HA^(-)] = y`, and `[A^(2-)] = z`.

The fraction of `[H_2A]` will be

`f_(H_2A) = x/(x + y + z)` (MAIN)

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

`x = ([H^(+)] * y)/K_(a1)` (A)

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

`y = ([H^(+)] * z)/K_(a2)` (B)

Plug (B) into (A)

`x = ([H^(+)] * [H^(+)] * z)/(K_(a1) * K_(a2)) = ([H^(+)]^(2) * z)/(K_(a1) * K_(a2))`

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

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

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

`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_2A) = ([H^(+)]^(2))/([H^(+)]^(2) + K_(a1) * [H^(+)] * K_(a1) * K_(a2))`

Answer 3

The speciation diagram is

Your equilibria are

H₂A ⇌ H⁺ + HA⁻; `"p"K_1 = 5.9`
HA⁻ ⇌ H⁺ + A²⁻; `"p"K_2 = 9.4`

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

The material balance is given by

`C = "[H₂A] + [HA⁻] + [A²⁻]"`, where `C` is the total concentration of the acid.

The fraction of each species is given by

`f_"H₂A" = "[H₂A]"/C`
`f_"HA⁻" = "[HA⁻]"/C`
`f_"A²⁻" = "[A²⁻]"/C`

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

Let `D = "[H⁺]"^2 + K_1"[H⁺]" + K_1K_2`

`f_"H₂A" = "[H⁺]"^2/D`; `f_"HA⁻" =( K_1"[H⁺]")/D`; `f_"A²⁻" = (K_1K_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 `"p"K_1` and `"p"K_2`.

The `color(purple)("purple")` HA⁻ line doesn't make it all the way back up to 1.00 because the `"p"K` values are so close.

But it does peak at pH 7.6, which is the average of `"p"K_1` and `"p"K_2`.