Chem – College: Acids and Bases that Have More than One Ka or Kb

What if acids or bases have more than one ionization?  More than one Ka or Kb?

The last practice problem from the last section points to what we will be exploring in this section. That is that you can sometimes get an acid or base that breaks into or creates more than one H+ or OH. Usually, when books and teachers talk about acids this way they call them polyprotic acids. This means that they release more than one proton, which is all the H+ is. As of yet I have not seen any words used to describe bases in the same way. All of this means that you can make more than one equilibrium equation for acids and bases that release or create more than one H+ or OH. The first release or creation of these ions is labeled number one. The second is number two and so on. Any example is below:

Examples:

Create the chemical equations and the Ka equilibrium equations for each ionization of the acid H2SO3.

H2SO3(aq) <—> H+(aq) + HSO3(aq)

 Ka1 = [ H+ ] [ HSO3– ] [ H2SO3 ]

HSO3(aq) <—> H+(aq) + SO32-(aq)

 Ka2 = [ H+ ] [ SO32- ] [ HSO3– ]

Create the chemical equations and the Kb equilibrium equations for each ionization of the base Be(OH)2.

Be(OH)2(aq) <—> OH(aq) + BeOH+(aq)

 Kb1 = [ BeOH+ ] [ OH– ] [ Be(OH)2 ]

BeOH+(aq) <—> OH(aq) + Be2+(aq)

 Kb2 = [ Be2+ ] [ OH– ] [ BeOH+ ]

VIDEO Setting up Polyionization Ka or Kb Equations Demonstrated Example 1: Create the chemical equations and the Kb equilibrium equations for each ionization of the base Se2-.

What is in the reactants of the first chemical equation we need to create?

Se2-(aq) + H2O(l) <—>

Se2- is acting as an acid or base? What does that mean it does?

Answer: It is acting as a base and therefore it grabs an H+ from the H2O to become HSe.

Se2-(aq) + H2O(l) <—> HSe(aq) + OH(aq)

What is the equilibrium equation for that chemical equation?

 Kb1 = [ HSe– ] [ OH– ] [ Se2- ]

How do you start the next chemical equation?

Answer: Use the product of the last chemical equation that had Se in it and add water.

HSe(aq) + H2O(l) <—>

HSe is acting as an acid or base? What does that mean it does?

Answer: It is acting as a base and therefore it grabs an H+ from the H2O to become H2Se.

HSe(aq) + H2O(l) <—> H2Se(aq) + OH(aq)

What is the equilibrium equation for that chemical equation?

 Kb2 = [ H2Se ] [ OH– ] [ HSe– ]

EVERYTHING IN RED IS THE ANSWER

One thing to keep in mind is with the vast majority of the time the first ionization (Ka1 or Kb1) is the only significant contributor to the equilibrium concentration. So, in our above example, most of the OH concentration comes from the creation of the HSe ion. This is a handy trick if you ever have to calculate the concentration of H+ or OH in an acid or base that has more than one ionization state. Really you only have to calculate the first concentration of H+ or OHbecause the second and third concentrations are so small they don’t even matter.

PRACTICE PROBLEMS: Create the chemical equations and the equilibrium equations from the information below.

Create the chemical equations and the Kb equilibrium equations for each ionization of the base CO32-.

CO32-(aq) + H2O(l) <—> HCO3(aq) + OH(aq)

 Kb1 = [ HCO3– ] [ OH– ] [ CO32- ]

HCO3(aq) + H2O(l) <—> H2CO3(aq) + OH(aq)

 Kb2 = [ H2CO3 ] [ OH– ] [ HCO3– ]

Create the chemical equations and the Ka equilibrium equations for each ionization of the acid H3PO4.

H3PO4(aq) <—> H+(aq) + H2PO4(aq)

 Ka1 = [ H+ ] [ H2PO4– ] [ H3PO4 ]

H2PO4(aq) <—> H+(aq) + HPO42-(aq)

 Ka2 = [ H+ ] [ HPO42- ] [ H2PO4– ]

HPO42-(aq) <—> H+(aq) + PO43-(aq)

 Ka3 = [ H+ ] [ PO43- ] [ HPO42- ]