CHM1 17 Polyprotic Acids Collection

Polyprotic acids have more than one proton to donate to water, and so they have more than one ionization constant (K_a1, K_a2, etc) that can be considered. Polyprotic bases take more than one proton from water, and also have more than one ionization constant (K_b1, K_b2, etc).

Most often the first proton exchange is the only one that considerably affects pH. This is discussed more at the end of the first example.

Example 6 (Polyprotic Acid):
What is the pH of a grapefruit that contains 0.007 M citric acid solution (C₆H₈O₇)?
(K_a1 = 7.5 x 10^-4, K_a2 = 1.7 x 10^-5, K_a3 = 4.0 x 10^-7)
 

Make an ICE table for the first dissociation

C₆H₈O₇(aq) + H₂O(l)

C₆H₇O₇^- + H₃O⁺(aq)

	C6H8O7	H2O	C6H7O7-	H3O+
I	0.007	---	0	0
C	-x	---	+x	+x
E	0.007 - x	---	x	x

and use the quadratic formula to find that
x = 0.00195 M = [H₃O⁺]

Then a second ICE table can be made for the second dissociation

C₆H₇O₇^-(aq) + H₂O(l)

C₆H₆O₇^2- + H₃O⁺(aq)

	C6H7O7-	H2O	C6H6O72-	H3O+
I	0.00195	---	0	0.00195
C	-x	---	+x	+x
E	0.00195 - x	---	x	0.00195 + x

Remember that, for the first dissociation, x = [H₃O⁺] = [C₆H₇O₇^-], so you can plug in the first value of x in for the initial concentrations of C₆H₇O₇^- and H₃O⁺.

and use the quadratic formula to find that
x = 1.67x10^-5
[H₃O⁺] = 0.00195 + 1.67x10^-5 = 0.00197 M
-log(0.00197) = pH = 2.71

Note that if you ignored the addition of hydronium from the second dissociation, then [H₃O⁺] = 0.00195 M, and using this value to calculate pH still gives you the answer of 2.71.

So even though you made two ICE tables (you could even make a third table for K_a3), the protons donated in the second dissociation were negligible compared to the first dissociation. So you can see that it is really only the first dissociation that affects pH. Most often this is the case, and only one ICE table is necessary. It is up to you how certain you want to be and how many ICE tables you want to make when you calculate these problems.

Example 7 (Polyprotic Base):
What is the pH of a saturated solution of sodium carbonate (Na₂CO₃)?
(solubility in water is 21.6g/100mL at room temperature)
(for carbonic acid, H₂CO₃, K_a1 = 4.5 x 10^-7, K_a2 = 4.7 x 10^-11

First, you have to find the find the initial concentration of CO₃^2- which can be found from
= 0.204 mol Na₂CO₃ = 0.204 mol CO₃^2-
then divide 0.204 mol by 0.100 L to get 2.04 M CO₃^2-

Plug into an ICE table

CO₃^2-(aq) + H₂O(l)

HCO₃^-(aq) + OH^-(aq)

	CO32-	H2O	HCO3-	OH-
I	2.04	---	0	0
C	-x	---	+x	+x
E	2.04 - x	---	x	x

But notice that the equilibrium constants are for carbonic acid.
If you were considering the dissociation of carbonic acid, you would write the following expressions

for H₂CO₃ + H₂O HCO₃^- + H₃O⁺
for HCO₃^- + H₂O CO₃^2- + H₃O⁺

The second acid ionization constant corresponds to the first base ionization constant (because the base reactions go backwards). To convert the second acid ionization constant to the first base ionization constant, you use the equation
K_a x K_b = K_w = 10^-14
so that
K_a2 x K_b1 = 10^-14
K_b1 = = 2.13 x 10^-4

Use the same equation to convert the first acid ionization constant to the second base ionization constant
K_a1 x K_b2 = 10^-14
K_b2 = = 2.22 x 10^-8

The expressions for the protonation of carbonate are now known to be

for CO₃^2- + H₂O HCO₃^- + OH^-
for HCO₃^- + H₂O H₂CO₃ + OH^-

Plug the ICE tables values into the first equilibrium expression

and use the quadratic formula to solve
x = 0.0207 M = [OH^-]

You can ignore the second base ionization constant because it removes a negligible amount of protons from water. If you want to test this by making an ICE table, you should get that the total hydroxide concentration is 0.0207000222 M 0.0207 M

so pOH = -log(0.0207) = 1.68

pH + pOH = 14
14 - 1.68 = pH = 12.32