Question #8298c

3 Answers
Mar 27, 2015

!! LONG ANSWER !!

So, you have your diprotic acid, H_2AH2A, and the equilibrium reactions that take place in aqueous solution

H_2A_((aq)) rightleftharpoons H_((aq))^(+) + HA_((aq))^(-)H2A(aq)H+(aq)+HA(aq), pK_(a1) = 5.9pKa1=5.9,

HA_((aq))^(-) rightleftharpoons H_((aq))^(+) + A_((aq))^(2-)HA(aq)H+(aq)+A2(aq), pK_(a2) = 9.4pKa2=9.4

This means that your solution will contain three species: H_2AH2A, HA^(-)HA, and A^(2-)A2. 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)[H2A][H2A]+[HA]+[A2]=fractionH2A

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)Ka1 and K_(a2)Ka2, you can write the fraction each species has by using the concentration of H^(+)H+, and K_(a1)Ka1 and K_(a2)Ka2.

These fractions will look like this

([H^(+)]^(2))/([H^(+)]^(2+) + K_(a1)*[H^(+)] + K_(a1) * K_(a2))[H+]2[H+]2++Ka1[H+]+Ka1Ka2 -> fraction of H_2AH2A

(K_(a1) * [H^(+)])/([H^(+)]^(2+) + K_(a1)*[H^(+)] + K_(a1) * K_(a2))Ka1[H+][H+]2++Ka1[H+]+Ka1Ka2 -> fraction of HA^(-)HA

(K_(a1) * K_(a2))/([H^(+)]^(2+) + K_(a1)*[H^(+)] + K_(a1) * K_(a2))Ka1Ka2[H+]2++Ka1[H+]+Ka1Ka2 -> fraction of A^(2-)A2

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^(+)][H+] by using

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

and the values of K_(a1)Ka1 and K_(a2)Ka2 from

K_(a1) = 10^(-pK_(a1))Ka1=10pKa1 and K_(a2) = 10^(-pK_(a2))Ka2=10pKa2

For the entire pH range your diagram will look like this

http://www.skepticalscience.com/Mackie_OA_not_OK_post_8.htmlhttp://www.skepticalscience.com/Mackie_OA_not_OK_post_8.html

That speciation diagram belongs to H_2CO_3H2CO3. Here's how you'd adapt yours.

  • The color(green)("green")green curve will be [H_2A][H2A];
  • The color(red)("red")red curve will be [HA^(-)][HA];
  • The color(blue)("blue")blue curve will be [A^(2-)][A2];

Now, your color(green)("green")green curve will intersect your color(red)("red")red curve at pK_(a1) = 5.9pKa1=5.9, and your color(red)("red")red curve will intersect your color(blue)("blue")blue curve at pK_(a2) = 9.4pKa2=9.4.

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

pH_("intermediate") = (pK_(a1) + pK_(a2))/2 = (5.9 + 9.4)/2 = 7.65 ~= 7.7pHintermediate=pKa1+pKa22=5.9+9.42=7.657.7

Mar 27, 2015

Here's a sample calculation of you you can derive the fractions formulas that only use [H^(+)][H+], K_(a1)Ka1, and K_(a2)Ka2

According to the definitions of K_(a1)Ka1 and K_(a2)Ka2, you'll get

(1): K_(a1) = ([H^(+)] * [HA^(-)])/([H_2A])Ka1=[H+][HA][H2A], and

(2): K_(a2) = ([H^(+)] * [A^(2-)])/([HA^(-)])Ka2=[H+][A2][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))

Mar 27, 2015

The speciation diagram is

enter image source here

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

enter image source here

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.