Chemistry Workshop 2B
Rolf Unterleitner/ Jim Hollister
Diprotic Acids
With diprotic (or triprotic) acid, there is more than one equilibrium equation. In fact, there
is a Ka equilibrium equation for each of the acidic protons. Since there is more than one Ka
equation, there is more than one unknown. As you know from algebra, we need as many
independent equations as unknowns if we want to solve for each of the unknowns exactly.
(Independent equations are not just algebraic variations of each other; they are different
expressions. You can’t just manipulate one equation and say it is a new, independent equation.)
Here is how it is done:
Example: diprotic acid H2B:
The two equilibrium equations are:
H2B(aq) + H2O(l) !" HB-(aq) + H3O+(aq)
Ka1
HB-(aq) + H2O(l) !" B-2(aq) + H3O+(aq)
Ka2
The two equilibrium constant expression (law of mass action) equations for these are:
Ka1 = [HB-(aq)] [H3O+(aq)]/[H2B(aq)] ; Ka2 = [ B-2(aq) ] [H3O+(aq) ]/[ HB-(aq) ]
From this we can see that we have four unknowns (five unknowns when we include [OH-]),
We need three more equations. Another equation we can use is the self-ionization
(autoprotolysis) of water equation.
2H2O(l) !" -OH(aq) + H3O+(aq)
Kw
The law of mass action expression for this equation is:
Kw = [-OH(aq)] [ H3O+(aq)]= 1.0 X 10-14 (Kw is called the ion product constant of water)
Now we have three equations and five unknowns. We need two more equations.
Because the number of moles of B-2 does not change, although what it is bonded to does
change,(H2B(aq), HB-(aq), B-2(aq)), we can express the material (mass) balance equation:
e.g. If the initial conc. is 0.50M H2B(aq), then:
0.50M H2B(aq) = [H2B(aq)] + [HB-(aq) ] + [ B-2(aq)]
We can also write out an equation using the condition of electroneutrality (charge balance
equation). This condition states that in any solution, the number of moles of positive charge is
equal to the number of moles of negative charge.
e.g. [H3O+(aq)] = [HB-(aq)] + 2* [B-2(aq)] + [OH-]
*For every one mole of B-2, there are 2 moles of negative charge. For the electroneutrality
equation only, the coefficient in front of each concentration is the same as the positive value of
the charge of that species.
Most professors only ask you to write the different types of equations that would be used for a
"brute force" calculation, which would use as many independent equations as you have
unknowns; they don't usually ask you to solve them all for all unknowns. Instead, to actually
solve a polyprotic acid problem, you can assume the dominant equilibrium, the largest Ka,
determines the concentration of H30+(aq), and then, if asked, you can determine other species
concentrations based on the values determined from the dominant equilibrium.
See if you can use some of these ideas to solve a problem.
Example: Consider a 0.20M H3PO3(aq) solution. For phosphorous acid at 25.0°C, pKa1 =1.43
and pKa2 = 6.68 (there are only 2 acidic protons) (It turns out that HPO3-2(aq) is not acidic.
O
H O
P
O H
this H not acidic Don’t let it worry you. I am telling you so you can do this
problem. Also, don't mix up phosphorous acid, H3PO3(aq), with phosphoric acid, H3PO4(aq),
which does have three acidic protons.)
H
a)
List all the concentration unknowns (express as concentrations) in the solution.
b)
Write out three independent, equilibrium-constant expressions applicable to the solution.
c)
Write out the material balance condition for "phosphite."
d)
Write out the electroneutrality condition for the sol'n.
e)
Calculate the pH of the sol'n.
f)
Calculate the value of [HPO3-2(aq)] at equilibrium.
Answers:
a) [H3PO3(aq)], [H2PO3-(aq)], [HPO3-2(aq)], [H3O+(aq)], [OH-(aq)], [PO3-3(aq)]
b)
Ka1 = [H2PO3-(aq)] [H3O+(aq)]/ [H3PO3(aq)]
Ka2 = [HPO3-2(aq)] [H3O+(aq)] / [H2PO3-(aq)]
Kw = [-OH(aq)] [H3O+(aq)] = 1.0 X 10-14
c)
0.20M H3PO3(aq) = [H3PO3(aq)] + [H2PO3-(aq)] + [HPO3-2(aq)]
d)
[H3O+(aq)] = [H2PO3-(aq)] + 2 [HPO3-2(aq)] + [OH-(aq)]
e)
pH = 1.16 (You must use the quadratic equation to find x; x is too large for approximating.)
f)
[HPO3-2(aq)] = 2.1X10-7M ([HPO3-2(aq)] = Ka2 ! )